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.
Read and Learn More WBCHSE Solutions For Class 12 Maths
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
Conditions for the Applicability of a Binomial Distribution
- The experiment is performed for a finite and fixed number of trials.
- Each trial must give either a success or a failure.
- 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
- 3 Heads
- No head
- At least one head
- 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
- P(X≥1)
- 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
- Exactly 4 heads
- At least 1 head
- At most 4 heads.
Solution:
- 15/64
- 63/64
- 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
- Exactly 3 heads
- Not more than 4 heads
- At least 4 heads.
Solution:
- 15/128
- 193/512
- 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
- Exactly 5 successes
- At least 5 successes
- At most 5 successes.
Solution:
- 3/32
- 7/64
- 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
- Exactly 3 successes
- At least 2 successes
- At most 2 successes.
Solution:
- 8/81
- 11/27
- 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
- No success
- Exactly 6 successes
- At least 6 successes
- At most 6 successes.
Solution:
- (5/6)²
- 35/67
- 1/65
- (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
- None is defective,
- Exactly 2 are defective?
Solution:
- (9/10)5
- 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
- None will fuse after 6 months of use
- At least one will fuse after 6 months of use
- Not more than one will fuse after 6 months of use.
Solution:
- (19/20)5
- 1-(19/20)5
- (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
- Exactly 2 defective items,
- Not more than 2 defective items,
- At least 3 defective items.
Solution:
- 3/20 x (9/10)4
- 3/2 x (19/20)4
- 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
- None is white,
- All are white,
- At least one is white.
Solution:
- 81/256
- 1/256
- 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.