WBCHSE Solutions For Class12 Maths Binomial Distribution Formula in Probability with Examples

Binomial Distribution

Success And Failure In An Experiment: There are certain kinds of experiments that have two possible outcomes. One of these two outcomes is called a success, while the other is called a failure.

For example, in tossing a coin, we get either a head or a tail. If getting head is taken as a success then getting a tail is a failure.

Bernoulli’s Theorem Let there be n independent trials in an experiment and let the random variable X denote the number of successes in these trials.

Let the probability of getting a success in a single trial be p and that of getting a failure be q so that p + q = 1. Then,

P(X =r) = \({ }^n C_r \cdot p^r \cdot q^{(n-r)}.\)

Proof: Let us denote a success by S and a failure by F.

A number of ways of getting r successes in n trials = nCr.

∴ P(X=r) = \({ }^n C_r\) • {P(S) • P(S) …r times} x {P(F) • P(F)… (n – r) times}

= \({ }^n C_r\) (p • p • p …r times) x[q • q • q … (n-r) times]

= \({ }^n C_r \cdot p^r \cdot q^{(n-r)}\)

Hence, P(X = r) = \({ }^n C_r \cdot p^r \cdot q^{(n-r)}\)

Remark: We have

WBCHSE Solutions For Class12 Maths Binomial Distribution Formula in Probability with ExamplesConditions for the Applicability of a Binomial Distribution

  1. The experiment is performed for a finite and fixed number of trials.
  2. Each trial must give either a success or a failure.
  3. The probability of success in each trial is the same.

Binomial Distribution Solved Examples

Example 1. A coin is tossed 4 times. If X is the number of heads observed, find the probability distribution of X.
Solution:

When a coin is tossed, we have S = {H, T}.

P(getting a head) = 1/2, and

P(not getting a head) = (1-1/2) = 1/2

Let X be the random variable denoting the number of heads.

In 4 trials, we may get 0 or 1 or 2 or 3 or 4 heads.

So, X may assume the values 0,1, 2, 3, 4.

P(X = 0) = \({ }^4 C_0 \cdot\left(\frac{1}{2}\right)^0 \cdot\left(\frac{1}{2}\right)^{(4-0)}=\frac{1}{16}\)

P(X = 1) = \({ }^4 C_1 \cdot\left(\frac{1}{2}\right)^1 \cdot\left(\frac{1}{2}\right)^{(4-1)}=\frac{1}{4}\)

P(X = 2) = \({ }^4 C_2 \cdot\left(\frac{1}{2}\right)^2 \cdot\left(\frac{1}{2}\right)^{(4-2)}=\frac{3}{8}\)

P(X = 3) = \({ }^4 C_3 \cdot\left(\frac{1}{2}\right)^3 \cdot\left(\frac{1}{2}\right)^{(4-3)}=\frac{1}{4}\)

P(X = 4) = \({ }^4 C_4 \cdot\left(\frac{1}{2}\right)^4 \cdot\left(\frac{1}{2}\right)^{(4-4)}=\frac{1}{16}\)

Hence, the required probability distribution is given by

⇒ \(\left(\begin{array}{llllll}
X: & 0 & 1 & 2 & 3 & 4 \\
P(X): & \frac{1}{16} & \frac{1}{4} & \frac{3}{8} & \frac{1}{4} & \frac{1}{16}
\end{array}\right)\)

Example 2. Find the probability distribution of the number of suffixes in three tosses of a die.
Solution:

When a die is tossed, we have S = {1, 2, 3, 4, 5, 6}.

∴ P(getting a six) = 1/6 and P(not getting a six) = (1-1/6) = 5/6

Let X be the random variable denoting the number of sixes.

In 3 trials, the number of sixes may be 0 or 1 or 2 or 3.

So, X may assume the values 0, 1, 2, 3.

P(X=0) = \({ }^3 C_0 \cdot\left(\frac{1}{6}\right)^0 \cdot\left(\frac{5}{6}\right)^{(3-0)}=\frac{125}{216}\)

P(X=1 )= \({ }^3 C_1 \cdot\left(\frac{1}{6}\right)^1 \cdot\left(\frac{5}{6}\right)^{(3-1)}=\frac{25}{72}\)

P(X=2) = \({ }^3 C_2 \cdot\left(\frac{1}{6}\right)^2 \cdot\left(\frac{5}{6}\right)^{(3-2)}=\frac{5}{72}\)

P(X=3) = \({ }^3 C_3 \cdot\left(\frac{1}{6}\right)^3 \cdot\left(\frac{5}{6}\right)^{(3-3)}=\frac{1}{216}\)

The required probability distribution of X is given below:

⇒ \(\left(\begin{array}{lcccc}
X: & 0 & 1 & 2 & 3 \\
P(X): & \frac{125}{216} & \frac{25}{72} & \frac{5}{72} & \frac{1}{216}
\end{array}\right) .\)

Example 3. Find the probability distribution of the number of doublets in four throws of a pair of dice.
Solution:

When a pair of dice is thrown, there are 36 possible outcomes.

∴ n(S)= 36.

All possible doublets are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6).

P(getting a doublet) = \(\frac{6}{36}\) = \(\frac{1}{6}\), and

P(not getting a doublet) = (1-\(\frac{1}{6}\)) = \(\frac{5}{6}\)

Let X denote the number of doublets.

In 4 throws, we can have 0 or1 2 or 3, or 4 doublets

P(X=0) = \({ }^4 C_0 \cdot\left(\frac{1}{6}\right)^0 \cdot\left(\frac{5}{6}\right)^4=\frac{625}{1296}\)

P(X=1) = \({ }^4 C_1 \cdot\left(\frac{1}{6}\right)^1 \cdot\left(\frac{5}{6}\right)^3=\frac{125}{324}\)

P(X=2) = \({ }^4 C_2 \cdot\left(\frac{1}{6}\right)^2 \cdot\left(\frac{5}{6}\right)^2=\frac{25}{216}\)

P(X=3) = \({ }^4 C_3 \cdot\left(\frac{1}{6}\right)^3 \cdot\left(\frac{5}{6}\right)^1=\frac{5}{324}\)

P(X=4) = \({ }^4 C_4 \cdot\left(\frac{1}{6}\right)^4 \cdot\left(\frac{5}{6}\right)^0=\frac{1}{1296}\)

The required probability distribution is given below:

⇒ \(\left(\begin{array}{lccccc}
X: & 0 & 1 & 2 & 3 & 4 \\
P(X): & \frac{625}{1296} & \frac{125}{324} & \frac{25}{216} & \frac{5}{324} & \frac{1}{1296}
\end{array}\right) .\)

Example 4. An unbiased coin is tossed 6 times. Find, using the binomial distribution, the probability of getting at least 5 heads.
Solution:

In a single throw of a coin, we have S = {H, T}.

P(getting a head) = \(\frac{1}{2}\) and P(not getting a head) = (1-\(\frac{1}{2}\)) = \(\frac{1}{2}\)

∴ p = \(\frac{1}{2}\) and q = \(\frac{1}{2}\)

P(X=r) = \({ }^n C_r \cdot p^r \cdot q^{(n-r)}={ }^6 C_r \cdot\left(\frac{1}{2}\right)^r \cdot\left(\frac{1}{2}\right)^{(6-r)}={ }^6 C_r \cdot\left(\frac{1}{2}\right)^6\)

∴ P(getting at least 5 heads) = P(X≥5)

= P(X = 5) + P(X = 6)

= \({ }^6 C_5 \cdot\left(\frac{1}{2}\right)^6+{ }^6 C_6 \cdot\left(\frac{1}{2}\right)^6=\left(\frac{3}{32}+\frac{1}{64}\right)=\frac{7}{64}\)

Hence, the required probability is \(\frac{7}{64}\)

Example 5. An unbiased coin is tossed 8 times. Find, by using binomial distribution, the probability of getting at least 3 heads.
Solution:

In a single throw of a coin, we have S = {H, T}.

P(getting a head) = \(\frac{1}{2}\) and P(not getting a head) = (1-\(\frac{1}{2}\)) = \(\frac{1}{2}\)

∴ p = \(\frac{1}{2}\) and \(\frac{1}{2}\)

P(X=r) = \({ }^n C_r \cdot p^r \cdot q^{(n-r)}={ }^8 C_r \cdot\left(\frac{1}{2}\right)^r \cdot\left(\frac{1}{2}\right)^{(8-r)}={ }^8 C_r\left(\frac{1}{2}\right)^8\)

∴ P(getting at least 3 heads) = P(X≥3)

= 1 – [P(X = 0) + P(X = 1) + P(X = 2)]

= \(1-\left[{ }^8 C_0 \cdot\left(\frac{1}{2}\right)^8+{ }^8 C_1 \cdot\left(\frac{1}{2}\right)^8+{ }^8 C_2 \cdot\left(\frac{1}{2}\right)^8\right]\)

= \(1-\frac{1}{256} \cdot(1+8+28)=\left(1-\frac{37}{256}\right)=\frac{219}{256}\)

Hence, the required probability is \(\frac{219}{256}\).

Example 6. Six coins are tossed simultaneously. Find the probability of getting

  1. 3 Heads
  2. No head
  3. At least one head
  4. Not more than 3 heads.

Solution:

The experiment may be taken as throwing a single coin 6 times.

In a single throw of a coin, we have S = {H, T}.

P(getting a head) = \(\frac{1}{2}\) and P(not getting a head) = (1-\(\frac{1}{2}\)) = \(\frac{1}{2}\)

Let X be the random variable showing the number of heads.

P(X = r) = \({ }^n C_r \cdot p^r \cdot q^{(n-r)}={ }^6 C_r \cdot\left(\frac{1}{2}\right)^r \cdot\left(\frac{1}{2}\right)^{6-r}={ }^6 C_r \cdot\left(\frac{1}{2}\right)^6\)

1. P(getting 3 heads) = P(X = 3) = \(={ }^6 C_3 \cdot\left(\frac{1}{2}\right)^6=\frac{5}{16}\)

2. P(getting no head) = P(X = 0) = \({ }^6 C_0 \cdot\left(\frac{1}{2}\right)^6=\frac{1}{64}\)

3. P(getting at least 1 head)

= \(1-P(X=0)=1-\left[{ }^6 C_0 \cdot\left(\frac{1}{2}\right)^6\right]\)

= \(\left(1-\frac{1}{64}\right)=\frac{63}{64}\)

4. P(getting not more than 3 heads) = P(no head or 1 head or 2 heads or 3 heads)

= P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

= \({ }^6 C_0 \cdot\left(\frac{1}{2}\right)^6+{ }^6 C_1 \cdot\left(\frac{1}{2}\right)^6+{ }^6 C_2 \cdot\left(\frac{1}{2}\right)^6+{ }^6 C_3 \cdot\left(\frac{1}{2}\right)^6\)

= \(\left(\frac{1}{2}\right)^6 \cdot(1+6+15+20)=\left(\frac{1}{64} \times 42\right)=\frac{21}{32}\)

Example 7. A die is thrown 5 times. If getting an odd number is a success, find the probability of getting at least 4 successes.
Solution:

When a die is thrown, we have S = {1,2,3,4,5,6}.

∴ P(getting an odd number) = \(\frac{3}{6}\) = \(\frac{1}{2}\)

∴ P(a success) = \(\frac{1}{2}\), and P(not a success) = (1-\(\frac{1}{2}\)) = \(\frac{1}{2}\)

Let X be the random variable showing the number of successes.

P(X=r) = \(={ }^n C_r \cdot p^r \cdot q^{(n-r)}={ }^5 C_r \cdot\left(\frac{1}{2}\right)^r \cdot\left(\frac{1}{2}\right)^{(5-r)}={ }^5 C_r \cdot\left(\frac{1}{2}\right)^5\)

P(at least 4 successes) = P(4 successes or 5 successes)

= P(X = 4) + P(X = 5)

= \({ }^5 C_4 \cdot\left(\frac{1}{2}\right)^5+{ }^5 C_5 \cdot\left(\frac{1}{2}\right)^5=\left(\frac{5}{32}+\frac{1}{32}\right)=\frac{6}{32}=\frac{3}{16}\)

Example 8. In 4 throws with a pair of dice, what is the probability of throwing doublets at least twice?
Solution:

In a single throw of a pair of dice, the number of all possible outcomes is 36.

All doublets are (1,1), (2,2), (3,3), (4,4), (5,5), (6,6).

P(getting a doublet) = \(\frac{6}{36}\) = \(\frac{1}{6}\) and,

P(not getting a doublet) = (1-\(\frac{1}{6}\)) = \(\frac{5}{6}\)

Let X be the random variable denoting the number of doublets.

Then, P(X = r) = \({ }^n C_r \cdot p^r \cdot q^{(n-r)}={ }^4 C_r \cdot\left(\frac{1}{6}\right)^r \cdot\left(\frac{5}{6}\right)^{(4-r)} .\)

P(at least 2 doublets) = P(X = 2) + P(X = 3) + P(X = 4)

= \({ }^4 C_2 \cdot\left(\frac{1}{6}\right)^2 \cdot\left(\frac{5}{6}\right)^{(4-2)}+{ }^4 C_3 \cdot\left(\frac{1}{6}\right)^3 \cdot\left(\frac{5}{6}\right)^{(4-3)}+{ }^4 C_4 \cdot\left(\frac{1}{6}\right)^4 \cdot\left(\frac{5}{6}\right)^{(4-4)}\)

= \(6 \cdot\left(\frac{1}{6}\right)^2 \cdot\left(\frac{5}{6}\right)^2+4 \cdot\left(\frac{1}{6}\right)^3 \cdot\left(\frac{5}{6}\right)^1+\left(\frac{1}{6}\right)^4 \cdot\left(\frac{5}{6}\right)^0\)

= \(\left(\frac{25}{216}+\frac{5}{324}+\frac{1}{1296}\right)=\frac{171}{1296}\)

Example 9. The bulbs produced in a factory are supposed to contain 5% defective bulbs. What is the probability that a sample of 10 bulbs will contain no more than 2 defective bulbs?
Solution:

P(getting a defective bulb) = \(\frac{5}{100}\) = \(\frac{1}{20}\), and

P(getting a non-defective bulb) = (1-\(\frac{1}{20}\)) = \(\frac{19}{20}\), and

Then, p = \(\frac{1}{20}\) and q = \(\frac{19}{20}\)

Let X denote the number of defective bulbs.

P(X=r) = \({ }^n C_r \cdot p^r \cdot q^{(n-r)}\)=\({ }^{10} C_r \cdot\left(\frac{1}{20}\right)^r \cdot\left(\frac{19}{20}\right)^{(10-r)}\)

P(getting not more than 2 defective bulbs)

= P(X = 0 or X = 1 or X = 2)

= P(X = 0) + P(X = 1) + P(X = 2)

= \({ }^{10} C_0 \cdot\left(\frac{1}{20}\right)^0 \cdot\left(\frac{19}{20}\right)^{10}+{ }^{10} C_1 \cdot\left(\frac{1}{20}\right)^1 \cdot\left(\frac{19}{20}\right)^9+{ }^{10} C_2 \cdot\left(\frac{1}{20}\right)^2 \cdot\left(\frac{19}{20}\right)^8\)

= \(\left(\frac{19}{20}\right)^{10}+\frac{1}{2} \cdot\left(\frac{19}{20}\right)^9+\frac{9}{80} \cdot\left(\frac{19}{20}\right)^8=\left(\frac{19}{20}\right)^8 \cdot\left(\frac{149}{100}\right)\)

Let A = \(\left(\frac{19}{20}\right)^8 \cdot\left(\frac{149}{100}\right) \cdot\) Then,

log A = 8(log 19 – log 20) + log 149 – log 100

= 8(1.2788 – 1.3010) + 2.1732 – 2 = -0.0044 = 1.9956.

∴ A = antilog (1.9956) = 0.99.

Hence, the required probability = \(\frac{99}{100}\)

Example 10. If on average, out of 10 ships, one gets drowned then what is the probability that out of 5 ships at least 4 reach the shore safely?
Solution:

Probability of a ship to reach the shore safely = \(\frac{9}{10}\)

Probability that a ship gets drowned = (1-\(\frac{9}{10}\)) = \(\frac{1}{10}\)

Let X be the random variable showing the number of ships reaching the shore safely.

∴ P(at least 4 reaching safely)

= P(4 reaching safely or 5 reaching safely)

= P(4 reaching safely) + P(5 reaching safely)

= \({ }^5 C_4 \cdot\left(\frac{9}{10}\right)^4 \cdot\left(\frac{1}{10}\right)^{(5-4)}+{ }^5 C_5 \cdot\left(\frac{9}{10}\right)^5 \cdot\left(\frac{1}{10}\right)^0\)

= \(\frac{1}{2} \cdot\left(\frac{9}{10}\right)^4+\left(\frac{9}{10}\right)^5=\left(\frac{9}{10}\right)^4\left(\frac{1}{2}+\frac{9}{10}\right)=\frac{7}{5} \cdot\left(\frac{9}{10}\right)^4\)

A = \(\frac{7}{5} \cdot\left(\frac{9}{10}\right)^4=\frac{7 \times(9)^4}{5 \times(10)^4}\)

⇒ logA = log7 + 4xlog9-log5-4xloglO = (0.8451 + 4 x 0.9542 – 0.6990- 4) = -0.0371 = 1.9629

⇒ A = antilog (1.9629) = 0.9181.

Hence, the required probability is 0.9181.

Mean and Variance of a Binomial Distribution

Mean: If a random variable X assumes the values x1, x2,… ,xn with probabilities p1, p2, ……., pn respectively then the mean of X is defined by

∴ \(\mu=\sum_{i=1}^n x_i p_i .\)

To Find the Mean of a Binomial Distribution

For the binomial distribution P(X = r) = P(r) = \({ }^n C_r \cdot p^r \cdot q^{(n-r)}\), where r = 0,1,2,… ,n the mean, p, is given by

⇒ \(\mu =\sum_{r=0}^n r \cdot p(r)=\sum_{r=0}^n r \cdot{ }^n C_r \cdot p^r \cdot q^{(n-r)}\)

= \({ }^n C_1 \cdot p \cdot q^{n-1}+2 \cdot{ }^n C_2 \cdot p^2 \cdot q^{(n-2)}+\ldots+n \cdot{ }^n C_n \cdot p^n q^0\)

= \(1 \cdot n p \cdot q^{n-1}+n(n-1) \cdot p^2 \cdot q^{(n-2)}+\ldots+n p^n\)

= \(n p \cdot\left[{ }^{(n-1)} C_0 \cdot p^0 \cdot q^{(n-1)}+{ }^{(n-1)} C_1 \cdot p^1 \cdot q^{(n-2)}+\ldots+{ }^{(n-1)} C_{(n-1)} \cdot p^{n-1} \cdot q^0\right]\)

= \((n p) \cdot(q+p)^{n-1}=(n p)\) [because q+p=1]

Hence, the mean is given by μ= np.

The variance =σ² is given by

⇒ \(\sigma^2=\sum_{r=0}^n r^2 \cdot p(r)-(\text { mean })^2\)

= \(\sum_{r=0}^n r^2 \cdot{ }^n C_r \cdot p^r q^{(n-r)}-(n p)^2\) [because mean=np]

= \(\sum_{r=0}^n\{r+r(r-1)\} \cdot{ }^n C_r p^r q^{(n-r)}-(n p)^2\)

= \(\sum_{r=0}^n r \cdot{ }^n C_r p^r q^{(n-r)}+\sum_{r=0}^n r(r-1) \cdot{ }^n C_r \cdot p^r \cdot q^{(n-r)}-(n p)^2\)

= \(n p+\sum_{r=2}^n r(r-1) \cdot \frac{n(n-1)}{r(r-1)} \cdot{ }^{(n-2)} C_{(r-2)} \cdot p^2 \cdot p^{(r-2)} \cdot q^{(n-r)}-(n p)^2\)

= \(n p+n(n-1) \cdot p^2 \cdot\left(\sum_{r=2}^n \sum_{r=0}^n r \cdot{ }^n C_r \cdot p^r \cdot q^{(n-r)}=\text { mean }=n p\right]\)

= \(n p+n(n-1) p^2(q+p)^{(n-2)}-n^2 p^2\)

= \(n p+n(n-1) p^2-n^2 p^2\) [because q+p=1]

= \(n p-n p^2=n p(1-p)=n p q\).

Hence, variance = npq.

∴ standard deviation = √npq.

Recurrence Relation for a Binomial Distribution

We have

P(r) = \({ }^n C_r \cdot p^r \cdot q^{(n-r)} \text { and } P(r+1)={ }^n C_{(r+1)} \cdot p^{(r+1)} \cdot q^{n-r-1} \text {. }\)

∴ \(\frac{P(r+1)}{P(r)}=\frac{{ }^n C_{(r+1)} \cdot p^{(r+1)} \cdot q^{n-r-1}}{{ }^n C_r \cdot p^r \cdot q^{(n-r)}}\)

= \(\frac{(n !)}{(r+1) ! \cdot(n-r-1) !} \cdot \frac{(r) ! \cdot(n-r) !}{(n) !} \cdot \frac{p}{q}=\frac{(n-r)}{(r+1)} \cdot \frac{p}{q}\)

∴ P(r+1) = \(\frac{(n-r)}{(r+1)} \cdot \frac{p}{q} \cdot P(r)\)

Example 11. If X follows a binomial distribution with mean 3 and variance (3/2), find

  1. P(X≥1)
  2. P(X≤5).

Solution:

We know that mean = np and variance = npq.

∴ np = 3 and npq = \(\frac{3}{2}\)

⇒ 3q = \(\frac{3}{2}\)

⇒ q = \(\frac{1}{2}\)

Now, np = 3 and p = \(\frac{1}{2}\)

⇒ n X \(\frac{1}{2}\) = 3

⇒ n = 6.

So, the binomial distribution is given by

P(X=r)= \({ }^n C_r \cdot p^r \cdot q^{(n-r)}={ }^6 C_r \cdot\left(\frac{1}{2}\right)^r \cdot\left(\frac{1}{2}\right)^{(6-r)}={ }^6 C_r\left(\frac{1}{2}\right)^6 \cdot\)

1. P(X≥1) =1-P(X=0)

= \(1-{ }^6 C_0 \cdot\left(\frac{1}{2}\right)^6=\left(1-\frac{1}{64}\right)=\frac{63}{64}\)

2. P(X≤5) =1-P(X= 0)

= \(1-{ }^6 C_6\left(\frac{1}{2}\right)^6=\left(1-\frac{1}{64}\right)=\frac{63}{64}\)

Example 12. If X follows a binomial distribution with mean 4 and variance 2, find P(X≥5).
Solution:

We know that mean = np and variance = npq.

∴ np = 4 and npq = 2.

Now, np = 4 and npq= 2

⇒ 4q = 2 ⇒ q = \(\frac{1}{2}\)

Now,np = 4 and p = \(\frac{1}{2}\)

⇒ \(\frac{1}{2}\) n = 4 ⇒ n = 8.

So, the binomial distribution is given by

P(X= r) = \({ }^n C_r \cdot p^r \cdot q^{(n-r)}={ }^8 C_r \cdot\left(\frac{1}{2}\right)^r \cdot\left(\frac{1}{2}\right)^{(8-r)}={ }^8 C_r \cdot\left(\frac{1}{2}\right)^8\)

∴ P(X > 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X= 8)

= \({ }^8 C_5 \cdot\left(\frac{1}{2}\right)^8+{ }^8 C_6 \cdot\left(\frac{1}{2}\right)^8+{ }^8 C_7 \cdot\left(\frac{1}{2}\right)^8+{ }^8 C_8 \cdot\left(\frac{1}{2}\right)^8\)

= \(\left[{ }^8 C_3+{ }^8 C_2+{ }^8 C_1+1\right]\left(\frac{1}{2}\right)^8\)

= \((56+28+8+1) \cdot \frac{1}{256}=\frac{93}{256}\)

Example 13. Find the binomial distribution for which the mean and variance are 12 and 3 respectively.
Solution:

Let X be a binomial variate for which mean = 12 and variance = 3.

Then, np =12 and npq = 3

⇔ 12xq = 3 ⇒ q = \(\frac{1}{4}\)

∴ p = (1-q) = (1-\(\frac{1}{4}\)) = \(\frac{3}{4}\)

And, np =12 ⇔ n = \(\frac{12}{p}\) = 12 x \(\frac{4}{3}\) = 16.

Thus, n =16, p = \(\frac{3}{4}\) and q = \(\frac{1}{4}\)

Hence, the binomial distribution is given by

P(X=r) = \({ }^{16} C_r \cdot\left(\frac{3}{4}\right)^r \cdot\left(\frac{1}{4}\right)^{(16-r)}\) where r = 0, 1, 2, 3, …,15.

Example 14. If the sum of the mean and variance of a binomial distribution for 5 trials is 1.8, find the distribution.
Solution:

We know that

mean = np and variance = npq.

It is being given that n = 5 and mean + variance = 1.8.

∴ np + npq = 1.8, where n = 5 +

⇔ 5p + 5pq = 1.8

⇔ p + p(1 -p) = 0.36 [q = (1-p)]

⇔ p²-2p + 0.36 = 0

⇔100p² – 200p + 36 = 0

⇔ 25p² – 50p + 9 = 0

⇔ 25p² -45p -5p + 9 = 0

⇔ 5p(5p – 9) – (5p – 9) = 0

⇔ (5p – 9)(5p -1) = 0 1

⇔ p = \(\frac{1}{5}\) = 0.2 [p cannot exceed 1].

Thus, n = 5, p = 0.2, and q = (1-p) = (1- 0.2) = 0.8.

Let X denote the binomial variate. Then, the required distribution is

P(X = r) = \({ }^n C_r \cdot p^r \cdot q^{(n-r)}={ }^5 C_r \cdot(0.2)^r \cdot(0.8)^{(5-r)} \text {, }\) where r = 0,1,2,3,4,5.

Example 15. The sum and the product of the mean and variance of a binomial distribution are 24 and 128 respectively. Find the distribution.
Solution:

We have np + npq – 24 and np x npq = 128

⇔ (np)( 1 + q) = 24 and n²p² x q = 128

⇔ n²p² = \(\frac{576}{(1+q)^2}\) and n²p² x q =128

⇔ \(\frac{576}{(1+q)^2}=\frac{128}{q}\)

⇔ 2(1 + q² + 2q) = 9q

⇔ 2q²-5q + 2 = 0

⇔ (2q – 1)(q – 2) = 0

⇔ q = \(\frac{1}{2}\) ⇔ q = 2

∴ p = (1-q) = (1-\(\frac{1}{2}\)) = \(\frac{1}{2}\)

Now, np(1+q) = 24

⇔ n x \(\frac{1}{2}\)(1+\(\frac{1}{2}\)) = 24

⇔ n = 32

Hence, the requied probability distribution is given by P(X=r) = \({ }^{32} C_r \cdot\left(\frac{1}{2}\right)^{32}\)

Example 16. In a binomial distribution, prove that mean > variance.
Solution:

Let X be a binomial variate with parameters n and p.

Then, mean = np and variance = npq.

∴ (mean)- (variance) = (np- npq) = np(1 -q) = np²> 0 [(1- q) = p and np²>0 as n∈N]

⇒ [(mean)- (variance)] > 0

⇒ mean > variance.

Hence, mean > variance.

Example 17. A die is tossed thrice. Getting an even number is considered a success. What is the variance of the binomial distribution?
Solution:

Here, n = 3.

Let p = probability of getting an even number in a single throw

⇒ p = \(\frac{3}{6}\) = \(\frac{1}{2}\)

⇒ q = (1-p) = (1-\(\frac{1}{2}\)) = \(\frac{1}{2}\)

∴ variance = npq = (3x\(\frac{1}{2}\)x\(\frac{1}{2}\)0 = \(\frac{3}{4}\)

Example 18. A die is rolled 20 times. Getting a number greater than 4 is a success. Find the mean and variance of the number of successes.
Solution:

In a single throw of a die, we have

p = probability of getting a number greater than 4 = \(\frac{2}{6}\) = \(\frac{1}{3}\)

∴ q = (1-p) = (1-\(\frac{1}{3}\)) = \(\frac{2}{3}\)

Also, n = 20 (given).

∴ mean = np =(20x\(\frac{1}{3}\))= 6.67, and

variance = npq =(20x\(\frac{1}{3}\)x\(\frac{2}{3}\))= 4.44.

Example 19. A die is tossed 180 times. Find the expected number (μ) of times the face with the number 5 will appear. Also, find the standard deviation (σ), and variance (σ²).
Solution:

In a single throw of a die, S = {1, 2, 3, 4, 5 6}.

p = (probability of getting the number 5) = \(\frac{1}{6}\)

∴ q = (1-p) = (1-\(\frac{1}{6}\)) = \(\frac{5}{6}\)

Thus, n = 180, p = \(\frac{1}{6}\) and q = \(\frac{5}{6}\)

∴ μ = np = (180x\(\frac{1}{6}\))=30.

Variance =  σ² = npq = (l80x\(\frac{1}{6}\)x\(\frac{5}{6}\)) = 25

Standard deviation = σ = √25 = 5

Binomial Distribution Exercise

Question 1. A coin is tossed 6 times. Find the probability of getting at least 3 heads.
Solution: 21/32

Hint: In a single toss, P(H) 1/2 and P(not H) = 1/2

⊂ p = 1/2, q = 1/2 and n = 6.

Let X show the number of heads. Then,

P(X= r) = \({ }^n C_r \cdot p^r \cdot q^{(n-r)}={ }^6 C_r\left(\frac{1}{2}\right)^6 .\)

Required probability = P(X= 3) +P(X= 4) +P(X= 5) +P(X= 6).

Question 2. A coin is tossed 5 times. What is the probability that a head appears an even number of times?
Solution: 1/2

Hint: Here,p = 1/2, q = 1/2 and n = 5

Let X show the number of heads. Then,

P(X=r) = \({ }^n C_r \cdot p^r \cdot q^{(n-n)}={ }^5 C_r\left(\frac{1}{2}\right)^5\).

Required probability=P(X= 0) + P(X= 2) + P(X= 4)

Question 3. 7 coins are tossed simultaneously. What is the probability that a tail appears an odd number of times?
Solution: 1/2

Hint: 7 coins being tossed simultaneously is the same as one coin being tossed 7 times.

∴ p = 1/2, q = 1/2 and n = 7.

Let X show the number of tails. Then,

P(X= r) = \({ }^n C_r \cdot p^r \cdot q^{n-r}={ }^7 C_r \cdot\left(\frac{1}{2}\right)^7\)

Required probability=P(X= 1) +P(X= 3) + P(X= 5) + P(X= 7).

Question 4. A coin is tossed 6 times. Find the probability of getting

  1. Exactly 4 heads
  2. At least 1 head
  3. At most 4 heads.

Solution:

  1. 15/64
  2. 63/64
  3. 57/64

Hint: P(a head) = 1/2 and P(not a head) = 1/2

∴ p = 1/2, q=2 and n = 6.

∴ P(X =r) = \({ }^n C_r \cdot p^r \cdot q^{(n-r)}={ }^6 C_r \cdot\left(\frac{1}{2}\right)^6 .\)

1. P(exactly 4 heads) = \({ }^6 C_4 \cdot\left(\frac{1}{2}\right)^6\)

2. P(at least1 head) =1- P(no head)

=1- P(0 head) = \(1-P(0 \text { head })=\left[1-{ }^6 C_0 \cdot\left(\frac{1}{2}\right)^6\right]\)

3. P(at the most 4 heads) = P(4 or less than 4 heads)

= 1-P[5 heads or 6 heads] = \(1-\left[\left({ }^6 C_5+{ }^6 C_6\right)\left(\frac{1}{2}\right)^6\right] .\)

Question 5. 10 coins are tossed simultaneously. Find the probability of getting

  1. Exactly 3 heads
  2. Not more than 4 heads
  3. At least 4 heads.

Solution:

  1. 15/128
  2. 193/512
  3. 53/64

Hint: 10 coins being tossed simultaneously is the same as one coin being tossed 10 times.

P(X=r) = \({ }^n C_r \cdot p^r \cdot q^{(n-r)}={ }^{10} C_r \cdot\left(\frac{1}{2}\right)^{10}\)

1. P(exactly 3 heads) = \({ }^{10} C_3 \cdot\left(\frac{1}{2}\right)^{10} .\)

2. P(not more than 4 heads)

= P(X ≤ 4)

= P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

= \(\left({ }^{10} C_0+{ }^{10} C_1+{ }^{10} C_2+{ }^{10} C_3+{ }^{10} C_4\right)\left(\frac{1}{2}\right)^{10}\)

3. P(at least 4 heads)

= P(4 heads or 5 heads or … or 10 heads)

= 1- P(0 head or 1 head or 2 heads or 3 heads)

= 1- [P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)]

= \(1-\left({ }^{10} C_0+{ }^{10} C_1+{ }^{10} C_2+{ }^{10} C_3\right)\left(\frac{1}{2}\right)^{10}\)

Question 6. A die is thrown 6 times. If getting an even number is a success, find the probability of getting

  1. Exactly 5 successes
  2. At least 5 successes
  3. At most 5 successes.

Solution:

  1. 3/32
  2. 7/64
  3. 63/64

Hint: p = \(\frac{3}{6}=\frac{1}{2}, q=\left(1-\frac{1}{2}\right)=\frac{1}{2} \text { and } n=6 . \)

P(X=r)= \({ }^n C_r \cdot p^r \cdot q^{n-r}={ }^6 C_5\left(\frac{1}{2}\right)^6\)

1. P(exactly 5 successes) = \({ }^n C_r \cdot p^r \cdot q^{n-r}={ }^6 C_5\left(\frac{1}{2}\right)^6\)

2. P(at least 5 successes) = P[(5 successes) or (6 successes)]

\(\left({ }^6 C_5+{ }^6 C_6\right)\left(\frac{1}{2}\right)^6\)

3. P(at most 5 successes) = P[0 or 1 or 2 or 3 or 5 successes]

= 1- P(6 successes) = \(\left[1-{ }^6 C_6 \cdot\left(\frac{1}{2}\right)^6\right]\)

Question 7. A die is thrown 4 times. “Getting a 1 or a 6” is considered a success. Find the probability of getting

  1. Exactly 3 successes
  2. At least 2 successes
  3. At most 2 successes.

Solution:

  1. 8/81
  2. 11/27
  3. 8/9

Hint: p= \(\frac{2}{6}=\frac{1}{3}, q=\left(1-\frac{1}{3}\right)=\frac{2}{3}\) and n=4.

P(X=r)= \({ }^n C_r \cdot p^r \cdot q^{(n-r)}={ }^4 C_r \cdot\left(\frac{1}{3}\right)^r \cdot\left(\frac{2}{3}\right)^{(4-r)}\).

1. P(exactly 3 successes)= \({ }^4 C_3 \cdot\left(\frac{1}{3}\right)^3 \cdot\left(\frac{2}{3}\right)^1\)

2. P(at least 2 successes) = [P(X = 2) or P(X = 3) or P(X = 4)] =1-[P(X = 0) + P(X=1)].

3. P(at most 2 successes) = P[(X = 0) or (X = l) or (X = 2)].

Question 8. Find the probability of a 4 turning up at least once in two tosses of a fair die.
Solution: 11/36

Hint: p= \(\frac{1}{6}, q=\left(1-\frac{1}{6}\right)=\frac{5}{6}\) and n=2.

P(X=r)= \({ }^n C_r \cdot p^r \cdot q^{(n-n)}={ }^2 C_r \cdot\left(\frac{1}{6}\right)^r \cdot\left(\frac{5}{6}\right)^{(2-r)}\).

P[at least one 4] = P(X=1 or X=2) = P(X =1) + P(X = 2)

Question 9. A pair of dice is thrown 4 times. If “getting a doublet’ is considered a success, find the probability of getting 2 successes.
Solution: 25/216

Hint: p= \(\frac{6}{36}=\frac{1}{6}, q=\left(1-\frac{1}{6}\right)=\frac{5}{6} \text { and } n=4 . \)

P(X=r)= \({ }^n C_r \cdot p^r \cdot q^{(n-r)}={ }^4 C_r \cdot\left(\frac{1}{6}\right)^r \cdot\left(\frac{5}{6}\right)^{(4-r)}\)

∴ P(X=2)= \({ }^4 C_2 \cdot\left(\frac{1}{6}\right)^2 \cdot\left(\frac{5}{6}\right)^2\).

Question 10. A pair of dice is thrown 7 times. If ‘getting a total of 7’ is considered a success, find the probability of getting

  1. No success
  2. Exactly 6 successes
  3. At least 6 successes
  4. At most 6 successes.

Solution:

  1. (5/6)²
  2. 35/67
  3. 1/65
  4. (1-1/67)

Hint: Let E={(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)}.

∴ p= \(\frac{6}{36}=\frac{1}{6}, q=\left(1-\frac{1}{6}\right)=\frac{5}{6}\), n=7

P(X=r)= \({ }^n C_r \cdot p^r \cdot q^{(n-r)}={ }^7 C_r \cdot\left(\frac{1}{6}\right)^r \cdot\left(\frac{5}{6}\right)^{(7-r)}\)

1. P(X=0)= \({ }^7 C_0 \cdot\left(\frac{5}{6}\right)^7\).

2. P(X=6)= \({ }^7 C_6 \cdot\left(\frac{1}{6}\right)^6 \cdot\left(\frac{5}{6}\right)\).

3. P(at least 6 successes)=P(6 successes or 7 successes)

= P(X=6)+P(X=7)

= \({ }^7 C_6 \cdot\left(\frac{1}{6}\right)^6 \cdot\left(\frac{5}{6}\right)+{ }^7 C_7 \cdot\left(\frac{1}{6}\right)^7\)

4. P(at most 6 successes)= P(X \≤ 6)

=[1-P(X=7)]= \(1-{ }^7 C_7 \cdot\left(\frac{1}{6}\right)^7\)

Question 11. There are 6% defective items in a large bulk of items. Find the probability that a sample of 8 items will include not more than one defective item.
Solution: (47/50)7x(71/50)

Hint: p= \(\frac{6}{100}=\frac{3}{50}, q=\left(1-\frac{3}{50}\right)=\frac{47}{50} \text { and } n=8 \text {. }\)

P(X=r)= \({ }^n C_r \cdot p^r \cdot q^{(n-r)}={ }^8 C_r \cdot\left(\frac{3}{50}\right)^r \cdot\left(\frac{47}{50}\right)^{(8-r)}\)

Required probability = P(0 defective or 1 defective)

=P(X=0)+P(X=1) = \({ }^8 C_0 \cdot\left(\frac{47}{50}\right)^8+{ }^8 C_1 \cdot\left(\frac{3}{50}\right)\left(\frac{47}{50}\right)^7 . \)

Question 12. In a box containing 60 bulbs, 6 are defective. What is the probability that out of a sample of 5 bulbs

  1. None is defective,
  2. Exactly 2 are defective?

Solution:

  1. (9/10)5
  2. 729/10000

Hint: p = \(\frac{6}{60}=\frac{1}{10}, q=\left(1-\frac{1}{10}\right)=\frac{9}{10}\) and n=5

P(X=r)= \({ }^n C_r \cdot p^r \cdot q^{n-r}={ }^5 C_r \cdot\left(\frac{1}{10}\right)^r \cdot\left(\frac{9}{10}\right)^{(5-r)}\)

1. P(none is defective)= \(P(X=0)={ }^5 C_0 \cdot\left(\frac{9}{10}\right)^5\).

2. P(exactly 2 are defective)= \(P(X=2)={ }^5 C_2 \cdot\left(\frac{1}{10}\right)^2 \cdot\left(\frac{9}{10}\right)^3\).

Question 13. The probability that a bulb produced by a factory will fuse after 6 months of use is 0.05. Find the probability that out of 5 such bulbs

  1. None will fuse after 6 months of use
  2. At least one will fuse after 6 months of use
  3. Not more than one will fuse after 6 months of use.

Solution:

  1. (19/20)5
  2. 1-(19/20)5
  3. (19/20)4x(6/5)

Hint: p= \(\frac{1}{20}, q=\left(1-\frac{1}{20}\right)=\frac{19}{20}\) and n=5.

P(X=r)= \({ }^n C_r \cdot p^r \cdot q^{(n-n)}={ }^5 C_r \cdot\left(\frac{1}{20}\right)^r \cdot\left(\frac{19}{20}\right)^{(5-r)}\)

1. P(X=0)= \({ }^5 C_0 \cdot\left(\frac{19}{20}\right)^5\).

2. \(P(X≥1)=1-P(X<1)=1-P(X=0)\)

= \(1-{ }^5 C_0 \cdot\left(\frac{19}{20}\right)^5\) \text

3. \(P(X \leq 1)=P(X=0)+P(X=1)\)

= \({ }^5 C_0 \cdot\left(\frac{19}{20}\right)^5+{ }^5 C_1 \cdot\left(\frac{1}{20}\right)\left(\frac{19}{20}\right)^4\)

Question 14. In the items produced by a factory, there are 10% defective items. A sample of 6 items is randomly chosen. Find the probability that this sample contains

  1. Exactly 2 defective items,
  2. Not more than 2 defective items,
  3. At least 3 defective items.

Solution:

  1. 3/20 x (9/10)4
  2. 3/2 x (19/20)4
  3. 1-[3/2x(19/20)4]

Hint: p= \(\frac{10}{100}\)=\(\frac{1}{10}\), q= \(\left(1-\frac{1}{10}\right)\)=\(\frac{9}{10}\) and n=6

P(X=r)= \({ }^n C_r \cdot p^r \cdot q^{(n-r)}={ }^6 C_r \cdot\left(\frac{1}{10}\right)^r \cdot\left(\frac{9}{10}\right)^{(6-n)}\)

1. P(X=2)= \({ }^6 C_2 \cdot\left(\frac{1}{10}\right)^2 \cdot\left(\frac{9}{10}\right)^4\).

2. \(P(X \leq 2)={ }^6 C_0 \cdot\left(\frac{9}{10}\right)^6+{ }^6 C_1 \cdot \frac{1}{10} \times\left(\frac{9}{10}\right)^5+{ }^6 C_2 \cdot \frac{1}{100} \times\left(\frac{9}{10}\right)^4\).

3. P(X≥3)=1-P(X<3)=1-P(X≤2)

Question 15. Assume that on average one telephone number out of 15, called between 3 p.m. and 4 p.m. on weekdays, will be busy. What is the probability that if six randomly selected telephone numbers are called, at least 3 of them will be busy?
Solution: 1-(14/15)4 .(59/45)

Hint: p= \(\frac{1}{15}, q=\left(1-\frac{1}{15}\right)=\frac{14}{15} \text { and } n=6\)

P(X=r)= \({ }^n C_r \cdot p^r \cdot q^{(n-n}={ }^6 C_r \cdot\left(\frac{1}{15}\right)^r \cdot\left(\frac{14}{15}\right)^{(6-n)} \text {. }\)

P(X≥3)=1-P(X<3)=1-[P(X=0)+P(X=1)+P(X=2)]

= \(1-\left[{ }^6 C_0 \cdot\left(\frac{14}{15}\right)^6+{ }^6 C_1 \cdot\left(\frac{1}{15}\right) \cdot\left(\frac{14}{15}\right)^5+{ }^6 C_2 \cdot\left(\frac{1}{15}\right)^2 \cdot\left(\frac{14}{15}\right)^4\right]\)

= \(1-\left(\frac{14}{15}\right)^4\left(\frac{59}{45}\right)\)

Question 16. Three cars participate in a race. The probability that any one of them has an accident is 0.1. Find the probability that all the cars reach the finishing line without any accident.
Solution: 729/1000

Hint: p= \(\frac{1}{10}, q=\left(1-\frac{1}{10}\right)=\frac{9}{10} \text { and } n=3 \)

P(X=r)= \({ }^n C_r \cdot p^r \cdot q^{(n-n)}={ }^3 C_r \cdot\left(\frac{1}{10}\right)^r \cdot\left(\frac{9}{10}\right)^{\beta-n}\)

P(X=0)= \({ }^3 C_0 \cdot\left(\frac{9}{10}\right)^3\)

Question 17. Past records show that 80% of the operations performed by a certain doctor were successful. If he performs 4 operations in a day, what is the probability that at least 3 operations will be successful?
Solution: 512/625

Hint: p= \(\frac{80}{100}=\frac{4}{5}, q=\left(1-\frac{4}{5}\right)=\frac{1}{5} \text { and } n=4 \text {. }\)

P(X=r)= \({ }^n C_r \cdot p^r \cdot q^{(n-n)}={ }^4 C_r \cdot\left(\frac{4}{5}\right)^r \cdot\left(\frac{1}{5}\right)^{(4-n)} \text {. }\)

P(X≥3)=P(X=3)+P(X=4)

= \(\left[{ }^4 C_3 \cdot\left(\frac{4}{5}\right)^3 \cdot\left(\frac{1}{5}\right)+{ }^4 C_4 \cdot\left(\frac{4}{5}\right)^4\right]\)

Question 18. The probability of a man hitting a target is (1/4). If he fires 7 times, what is the probability of his hitting the target at least twice?
Solution: 4547/8192

Hint: p=\(\frac{1}{4}, q=\left(1-\frac{1}{4}\right)=\frac{3}{4}\) and n=7

P(X=r)=\({ }^n C_r \cdot p^r \cdot q^{(n-n}={ }^7 C_r \cdot\left(\frac{1}{4}\right)^r \cdot\left(\frac{3}{4}\right)^{(n-n)}\)

P(X≥2)=1-P(X<2) =1-P[(X = 0) or (X=1)] =1-[P(X = 0) + P(X=1)].

Question 19. In a hurdles race, a player has to cross 10 hurdles. The probability that he will clear each hurdle is (5/6). What is the probability that he will knock down fewer than 2 hurdles?
Solution: \(\frac{5^{10}}{\left(2 \times 6^9\right)}\)

Hint: p=P(knocking down 1 hurdle)= \(\left(1-\frac{5}{6}\right)=\frac{1}{6}\), q=\(\frac{5}{6}\) and n=10.

P(X=r)= \({ }^n C_r \cdot p^r \cdot q^{(n-r)}={ }^{10} C_r \cdot\left(\frac{1}{6}\right)^r \cdot\left(\frac{5}{6}\right)^{(10-r)}\)

P(X<2)=P(X=0)+P(X=1)

= \(\left[{ }^{10} C_0 \cdot\left(\frac{5}{6}\right)^{10}+{ }^{10} C_1 \cdot\left(\frac{1}{6}\right)^1 \cdot\left(\frac{5}{9}\right)^9\right]\)

Question 20. A man can hit a bird, once in 3 shots. On this assumption he fires 3 shots. What is the chance that at least one bird is hit?
Solution: 19/27

Hint: p = \(\frac{1}{3}\), q= \(\left(1-\frac{1}{3}\right)=\frac{2}{3}\) and n=3.

P(X=r)= \({ }^n C_r \cdot p^r \cdot q^{(n-r)}={ }^3 C_r \cdot\left(\frac{1}{3}\right)^r \cdot\left(\frac{2}{3}\right)^{(3-r)} \cdot\)

⇒ \(P(X \geq 1) =P(X=1)+P(X=2)+P(X=3)\)

= \(\left[{ }^3 C_1 \times \frac{1}{3} \times\left(\frac{2}{3}\right)^2\right]+\left[{ }^3 C_2 \times\left(\frac{1}{3}\right)^2 \times \frac{2}{3}\right]+\left[{ }^3 C_3 \times\left(\frac{1}{3}\right)^3\right]\)

= \(\left(\frac{4}{9}+\frac{2}{9}+\frac{1}{9}\right)=\frac{7}{9}\)

Question 21. If the probability that a man aged 60 will live to be 70 is 0.65, what is the probability that out of 10 men, now 60, at least 8 will live to be 70?
Solution: 0.2615

Hint: \(P(X \geq 8)=P(X=8)+P(X=9)+P(X=10)\)

= \(\left[{ }^{10} C_8 \cdot\left(\frac{13}{20}\right)^8 \cdot\left(\frac{7}{20}\right)^2+{ }^{10} C_9 \cdot\left(\frac{13}{20}\right)^9 \cdot\left(\frac{7}{20}\right)+{ }^{10} C_{10} \cdot\left(\frac{13}{20}\right)^{10}\right]\)

= \(\left[{ }^{10} C_2 \cdot\left(\frac{13}{20}\right)^8 \cdot \frac{49}{400}+{ }^{10} C_1 \cdot\left(\frac{13}{20}\right)^9 \cdot \frac{7}{20}+\left(\frac{13}{20}\right)^{10}\right] \text {. }\)

∴ P=\(\left(\frac{13}{20}\right)^8 \cdot\left[\frac{441}{80}+\frac{91}{40}+\frac{169}{400}\right]=\left[\left(\frac{13}{20}\right)^8 \times \frac{821}{100}\right] \text {. }\)

⇒ \(\log P=[8(\log 13-\log 20)+\log 821-\log 100]\)

=[-1.4968+2.9143-2]=-0.5825 = \(\overline{1}\).4175

⇒ P= antilog \((\overline{1} .4175)\)

Question 22. A bag contains 5 white, 7 red, and 8 black balls. If four balls are drawn one by one with replacement, what is the probability that

  1. None is white,
  2. All are white,
  3. At least one is white.

Solution:

  1. 81/256
  2. 1/256
  3. 175/256

Hint:

P(white)=\(\frac{5}{20}=\frac{1}{4}, P(\text { non-white })=\frac{3}{4}\)

∴ p= \(\frac{1}{4}, q=\frac{3}{4} \text { and } n=4\)

P(X=r)=\({ }^n C_r \cdot p^r \cdot q^{(n-r)}={ }^4 C_r \cdot\left(\frac{1}{4}\right)^r \cdot\left(\frac{3}{4}\right)^{(4-r)}\)

1. P(X=0)= \({ }^4 C_0 \cdot\left(\frac{3}{4}\right)^4\).

2. P(X=4)=\({ }^4 C_4 \cdot\left(\frac{1}{4}\right)^4\).

3. \(P(X \geq 1)=1-P(X<1)=1-P(X=0)\).

Question 23. A policeman fires 6 bullets at a burglar. The probability that the burglar will be hit by a bullet is 0.6. What is the probability that the burglar is still unhurt?
Solution: 0.004096

Hint: p=\(\frac{6}{10}=\frac{3}{5}, q=\left(1-\frac{3}{5}\right)=\frac{2}{5}\) and n=6.

P(X=r)= \({ }^n C_r \cdot p^r \cdot q^{(n-r)}={ }^6 C_r \cdot\left(\frac{3}{5}\right)^r \cdot\left(\frac{2}{5}\right)^{(6-r)}\)

∴ P(X=0)= \({ }^6 C_0 \cdot\left(\frac{2}{5}\right)^6\)

Question 24. A die is tossed thrice. A success is 1 or 6 on a toss. Find the mean and variance of successes.
Solution: μ = 1, σ² = 2/3

Hint: p= \(\frac{2}{6}=\frac{1}{3}, q=\left(1-\frac{1}{3}\right)=\frac{2}{3} \text { and } n=3\)

∴ μ= np and σ²= npq

Question 25. A die is thrown 100 times. Getting an even number is considered a success. Find the mean and variance of sucess.
Solution: μ = 50, σ² = 25

Question 26. Determine the binomial distribution whose mean is 9 and variance is 6.
Solution: \({ }^{27} C_r \cdot\left(\frac{1}{3}\right)^r \cdot\left(\frac{2}{3}\right)^{(27-r)}, \text { where } r=0,1,2,3, \ldots, 27\)

Question 27. Find the binomial distribution whose mean is 5 and variance is 2.5.
Solution: \({ }^{10} C_r \cdot\left(\frac{1}{2}\right)^r \cdot\left(\frac{1}{2}\right)^{(10-r)}, 0 \leq r \leq 10\)

Question 28. The mean and variance of a binomial distribution are 4 and (4/3) respectively. Find P(X≥1).
Solution: 728/729

Hint: \(\left(n p=4 \text { and } n p q=\frac{4}{3}\right) \Rightarrow q=\frac{1}{3}\)

∴ p=(1-q)=\(\left(1-\frac{1}{3}\right)=\frac{2}{3}\)

and \(n=\frac{4}{p}=\left(4 \times \frac{3}{2}\right)=6 \)

∴ P(X=r)=\({ }^n C_r \cdot p^r \cdot q^{(n-r)}\)

= \(P(X \geq 1)=1-P(X=0) \cdot\left(\frac{2}{3}\right)^r \cdot\left(\frac{1}{3}\right)^{(6-r)}\)

= \(1-{ }^6 C_0 \cdot\left(\frac{2}{3}\right)^0 \cdot\left(\frac{1}{3}\right)^6\)

= \(\left(1-\frac{1}{3^6}\right)=\frac{728}{729}\)

Question 29. For a binomial distribution, the mean is 6 and the standard deviation is √2 ‘. Find the probability of getting 5 successes.
Solution: \({ }^9 C_5 \cdot\left(\frac{2}{3}\right)^5 \cdot\left(\frac{1}{3}\right)^4\)

Hint: np = 6 and npq = 2

⇒ q = \(\frac{1}{3}\) and p = \(\frac{2}{3}\)

⇒ \(\left(n \times \frac{2}{3}\right)=6 \Rightarrow n=\left(6 \times \frac{3}{2}\right)=9\)

∴ \(P(5 \text { successes })={ }^9 C_5 \cdot\left(\frac{2}{3}\right)^5 \cdot\left(\frac{1}{3}\right)^4\)

Question 30. In a binomial distribution, the sum and the product of the mean and the variance are (25/3) and (50/3) respectively. Find the distribution.
Solution: \({ }^{15} C_r \cdot\left(\frac{1}{3}\right)^r \cdot\left(\frac{2}{3}\right)^{(15-r)}\)

Question 31. Obtain the binomial distribution whose mean is 10 and the standard deviation is 2√2.
Solution: \({ }^{50} C_r \cdot\left(\frac{1}{5}\right)^r \cdot\left(\frac{4}{5}\right)^{(50-r)}, 0 \leq r \leq 50\)

Question 32. Bring out the fallacy, if any, in the following statement: ‘The mean of a binomial distribution is 6 and its variance is 9’.
Solution: The probability of getting a failure (i.e., q) cannot be greater than1

Hint: np = 6 and npq = 9

⇒ q = \(\frac{n p q}{n p}=\frac{9}{6}=\frac{3}{2}\)

But, q cannot be greater than 1.

 

 

 

 

 

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