Important Questions For CBSE Class 12 Maths Chapter 13 Probability

CBSE Class 12 Maths Chapter 13 Probability Important Questions

Question 1. If A and B are two independent events and P(A) = \(\frac{1}{3}\), P(B) = \(\frac{1}{2}\); find \(\mathrm{P}(\overline{\mathrm{A}} \mid \overline{\mathrm{B}})\).

Solution:

Given, A and B are two independent events and P(A) = \(\frac{1}{3}\), P(B) = \(\frac{1}{2}\)

⇒ P(A∩B) = P(A).P(B) = \(\frac{1}{3}\) x \(\frac{1}{2}\) = \(\frac{1}{6}\)

⇒ P(A/B) = \(\frac{P(A \cap B)}{P(B)}=\frac{\frac{6}{1}}{\frac{1}{2}}=\frac{1}{3}\)

⇒ \(\mathrm{P}(\overline{\mathrm{A}} / \overline{\mathrm{B}})=\mathrm{P}(\overline{\mathrm{A} / \mathrm{B}})=1-\mathrm{P}(\mathrm{A} / \mathrm{B}) \Rightarrow \mathrm{P}(\overline{\mathrm{A}} / \overline{\mathrm{B}})=1-\frac{1}{3}=\frac{2}{3}\)

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Question 2. A coin is tossed once. If the head comes up, a die is thrown; blit if the tail comes up, the coin is tossed again. Find the probability of obtaining a head and number 6.

Solution:

S = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, H), (T, T)}

Let event A denote “obtaining head” and B denote “obtaining number 6”

⇒ P(A) = \(\frac{1}{2}\), P(B) = \(\frac{1}{6}\)

Hence, the probability of obtaining head and number 6

= P[{(H, 6)}]= P(A∩B) = P(A).P(B) = \(\frac{1}{2}\) x \(\frac{1}{6}\) = \(\frac{1}{12}\)

Probability Class 12 Important Questions

Question 3. Two cards are drawn successively with replacement from a well-shuffled pack of 52 cards. Find the probability distribution of the number of spade cards.

Solution:

Let random variable X denote the number of spade cards; then the possible values of X are 0, 1 or 2.

P(X = 0) = P(no spade and no spade) = \(\frac{39}{52} \times \frac{39}{52}=\frac{3}{4} \times \frac{3}{4}=\frac{9}{16}\)

P(X = 1) = P(spade and no spade or no spade and spade)

= \(\left(\frac{13}{52} \times \frac{39}{52}\right)+\left(\frac{39}{52} \times \frac{13}{52}\right)=\left(\frac{1}{4} \times \frac{3}{4}\right)+\left(\frac{3}{4} \times \frac{1}{4}\right)=\frac{3}{8}\)

P(X = 2) = P(spade and spade) = \(\left(\frac{13}{52} \times \frac{13}{52}\right)=\left(\frac{1}{4} \times \frac{1}{4}\right)=\frac{1}{16}\)

Hence, the probability distribution of X is:

Question 4. A pair of dice is thrown and the sum of the numbers appearing on the dice is observed to be 7. Find the probability that the number 5 has appeared on at least one die.

Or,

The probability that A hits the target is \(\frac{1}{3}\) and the probability that B hits it is \(\frac{2}{5}\). If both try to hit the target independently, find the probability that the target is hit.

Solution:

When a pair of dice is thrown, the sample space S contains 36 outcomes.

Let E: Event that number 5 has appeared on at least one die.

F: Event that the sum of the numbers on the dice is 7

⇒ E = {(5,1), (5,2), (5,3), (5, 4), (5, 5), (5,6), (1,5), (2,5), (3,5), (4,5), (6,5)}, F = {(1, 6), (2,5), (3,4), (4,3), (5,2), (6,1)}

∴ E ∩ F = {(5, 2), (2,5)}

Probability Class 12 Important Questions

Now; P(E∩F) = \(\frac{2}{36}\),P(F) = \(\frac{6}{36}\)

Hence, required probability = \(P(E / F)=\frac{P(E \cap F)}{P(F)}=\frac{(2 / 36)}{(6 / 36)}=\frac{1}{3}\)

Given; P(E₁) = \(\frac{1}{3}\), P(E2) =\(\frac{2}{5}\)

⇒ \(P\left(\overline{E_1}\right)=\frac{2}{3} \cdot P\left(\overline{E_2}\right)=\frac{3}{5}\)

⇒ Required Probability = P(target is hit)

= \(1-\mathrm{P}\left(\overline{\mathrm{E}}_1 \overline{\mathrm{E}}_2\right)=1-\mathrm{P}\left(\overline{\mathrm{E}}_1\right) \cdot \mathrm{P}\left(\overline{\mathrm{E}}_2\right)=1-\left(\frac{2}{3} \times \frac{3}{5}\right)=1-\frac{2}{5}=\frac{3}{5}\)

Or,

Let E₁: Event that A hits the target

Let E₂: Event that B hits the target

Given; \(P\left(E_1\right)=\frac{1}{3}, P\left(E_2\right)=\frac{2}{5} \Rightarrow P\left(\overline{E_1}\right)=\frac{2}{3} \cdot P\left(\overline{E_2}\right)=\frac{3}{5}\)

⇒ Required Probability = P (target is hit)

= \(1-\mathrm{P}\left(\overline{\mathrm{E}}_1 \overline{\mathrm{E}}_2\right)=1-\mathrm{P}\left(\overline{\mathrm{E}}_1\right) \cdot \mathrm{P}\left(\overline{\mathrm{E}}_2\right)=1-\left(\frac{2}{3} \times \frac{3}{5}\right)=1-\frac{2}{5}=\frac{3}{5}\)

Question 5. A random variable X has the probability distribution:

Find the value of K and P(X ≤ 2).

Solution:

We know that ∑ P(X) = 1 (for probability distribution)

⇒ 0 + K + 4K + 3K + 2K = 1

⇒ 10 K = 1

⇒ K = \(\frac{1}{10}\)…..(1)

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

= 0 + k + 4K = 5K = \(\frac{5}{10}\) = \(\frac{1}{2}\) (from (1))

Probability Class 12 Important Questions

Question 6. A purse contains 3 silver and 6 copper coins and a second purse contains 4 silver and 3 copper coins. If a coin is drawn at random from one of the two purses, find the probability that it is a silver coin.

Solution:

Let E₁: Event that first purse is selected. E₂: Event that second purse is selected and A: Event that silver coin is drawn

⇒ \(P\left(E_1\right)=\frac{1}{2}, P\left(E_2\right)=\frac{1}{2}, P\left(A / E_1\right)=\frac{3}{9}, P\left(A / E_2\right)=\frac{4}{7}\)

∴ \(P(A)-P\left(E_1\right) P\left(A / E_1\right)+P\left(E_2\right) \cdot P\left(A / E_2\right)\) (by total probability theorem)

= \(\left(\frac{1}{2} \times \frac{3}{9}\right)+\left(\frac{1}{2} \times \frac{4}{7}\right)=\frac{1}{6}+\frac{2}{7}=\frac{19}{42}\)

Question 7. A coin is tossed 5 times. What is the probability of getting

1. 3 heads,
2. At most 3 heads?

Or,

Find the probability distribution of X, the number of heads in a simultaneous loss of two coins.

Solution:

1. P 3 Heads= \({ }^5 \mathrm{C}_3\left(\frac{1}{2}\right)^3\left(\frac{1}{2}\right)^2=10 \times \frac{1}{8} \times \frac{1}{4}=\frac{5}{16}\)

2. P (at most 3 Heads) =\(1-\mathrm{P}(4)-\mathrm{P}(5)\)

= \(1-{ }^3 C_4\left(\frac{1}{2}\right)^4\left(\frac{1}{2}\right)-{ }^3 C_5\left(\frac{1}{2}\right)^5=1-\frac{5}{32}-\frac{1}{32}=\frac{26}{32}=\frac{13}{16}\)

Or,

Let X denotes the number of heads in a simultaneous toss of two coins, then the possible values of X are 0, 1 or 2.

P(X=0)=P(TT)= \(\left(\frac{1}{2} \cdot \frac{1}{2}\right)=\frac{1}{4}\),

P(X=1)=P(HT)+P(T H)=\(\left(\frac{1}{2} \cdot \frac{1}{2}\right)+\left(\frac{1}{2} \cdot \frac{1}{2}\right)=\frac{2}{4}\),

P(X=2)=P(HH)=\(\left(\frac{1}{2} \cdot \frac{1}{2}\right)=\frac{1}{4}\)

Hence; the probability distribution of X is given as :

Question 8. If P(not A) = 0.7, P(B) = 0.7 and P(B/A) = 0.5, then find P( A/B)

Solution:

Given; \(\mathrm{P}(\overline{\mathrm{A}})\) = 0.7 ⇒ 1 – P(A) = 0.7 ⇒ P(A) = 0.3 and P(B) = 0.7

Also, P(B/A) = 0.5; P(A/B) = ?

We know that: P(B/A) = \(\frac{P(B \cap A)}{P(A)} \Rightarrow 0.5=\frac{P(B \cap A)}{0.3}\)

⇒ P(B ∩ A) = 0.5 x 0.3 = 0.15

Now, P(A ∩ B) = P(B ∩ A) = 0.15

⇒ P(A/B) = \(\frac{P(A \cap B)}{P(B)}=\frac{0.15}{0.70}=\frac{15}{70}=\frac{3}{14}\)

Question 9. A black and a red die are rolled together. Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.

Solution:

Let E: Event of obtaining the sum 8 on the dice.

F: Event that red die resulted in a number less than 4, and let first and second die represent the black and red die respectively.

⇒ E = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}

⇒ n(E) = 5,

F = {(1,1),(1,2),(1,3),(2,1),(2,2),(2,3)}(3,1),(3,2),(3,3),(4,1),(4,2),(4,3),(5,1),(5,2),(5,3),(6,1),(6,2),(6,3)}

⇒ n(F) = 18

and E ∩ F = {(5, 3), (6, 2)}

⇒ n(E ∩F) = 2

Here, \(\mathrm{P}(\mathrm{F})=\frac{18}{36}=\frac{1}{2}$ and $\mathrm{P}(\mathrm{E} \cap \mathrm{F})=\frac{2}{36}=\frac{1}{18}\)

∴ Required probability = \(P(E / F)=\frac{P(E \cap F)}{P(F)}=\frac{1 / 18}{1 / 2}=\frac{1}{9}\)

Probability Class 12 Important Questions

Question 10. A shopkeeper sells three types of flower seeds A₁, A₂, A₃. They are sold in the form of a mixture, where the proportions of these seeds are 4:4:2 respectively. The germination rates of the three types of seeds are 45%, 60% and 35% respectively. Based on the above information, answer the following questions

1. Calculate the probability that a randomly chosen seed will germinate.
2. Calculate the probability that the seed is of type A,, given that a randomly chosen seed germinates.

Solution:

Given, A₁: A₂: A₃ = 4 : 4 : 2

⇒ \(P\left(A_1\right)=\frac{4}{10}, P\left(A_2\right)=\frac{4}{10}\) and \(P\left(A_3\right)=\frac{2}{10}\)

Let E be the event that a seed germinates.

∴ \(P\left(\frac{E}{A_1}\right)=\frac{45}{100} \cdot P\left(\frac{E}{A_2}\right)=\frac{60}{100}\) and \(P\left(\frac{E}{A_7}\right)=\frac{35}{100}\)

1. \(\mathrm{P}(\mathrm{E})=\mathrm{P}\left(\mathrm{A}_1\right) \cdot \mathrm{P}\left(\mathrm{E} / \mathrm{A}_1\right)+\mathrm{P}\left(\mathrm{A}_2\right) \cdot \mathrm{P}\left(\mathrm{E} / \mathrm{A}_2\right)+\mathrm{P}\left(\mathrm{A}_3\right) \cdot \mathrm{P}\left(\mathrm{E} / \mathrm{A}_3\right)\)….(1)

= \(\left(\frac{4}{10} \cdot \frac{45}{100}\right)+\left(\frac{4}{10} \cdot \frac{60}{100}\right)+\left(\frac{2}{10} \cdot \frac{35}{100}\right)=\frac{490}{1000}=\frac{49}{100}=0.49\)

2. \(P\left(A_2 / E\right)=\frac{P\left(A_2\right) \cdot P\left(E / A_2\right)}{P\left(A_1\right) \cdot P\left(E / A_1\right)+P\left(A_2\right) \cdot P\left(E / A_2\right)+P\left(A_3\right) \cdot P\left(E / A_3\right)}\)

= \(\left(\frac{\frac{4}{10} \times \frac{60}{100}}{\frac{49}{100}}\right)\) (from (1))

= \(\frac{24}{49}=0.48\)

Question 11. A student either knows or guesses or copies the answer to a multiple-choice question with four choices. The probability that he makes a guess is 1/3 and the probability that he copies the answer is 1/6. The probability that his answer is correct given that he copied, is 1/8. Let E₁, E₂ and E₃ are the events that the student guesses, copies or knows the answer respectively and A is the event that the student answers correctly.

1. What is the probability that he answers correctly, given that he knows the answer?

1. 1
2. 1/2
3. 2/3
4. 1/4

2. What is the probability that he answers correctly, given that he knows the answer?

1. 1
2. 0
3. 1/4
4. 1/8

3. What is the probability that he answers correctly, given that he had made a guess?

1. 1/4
2. 0
3. 1
4. 1/8

4. What is the probability that he knew the answer to the question, given that he answered it correctly?

1. 24/29
2. 4/29
3. 1/29
4. 3/29

5. \(\sum_{k=1}^3 P\left(E_1 \mid A\right) \text { is : }\)

1. 0
2. 1/3
3. 1
4. 11/8

Solution:

Given that:

E₁ = Event that the student guesses the answer.

E₂ = Event that the student copies the answer.

E₃ = Event that the student knows the answer,

and A = event that the student answers correctly

Now, \(P\left(E_1\right)=\frac{1}{3}\) and \(P\left(E_2\right)=\frac{1}{6}\)

∴ \(P\left(E_1\right)+P\left(E_2\right)+P\left(E_3\right)=1 \Rightarrow \frac{1}{3}+\frac{1}{6}+P\left(E_3\right)=1\)

or \(P\left(E_3\right)=1-\frac{3}{6}=\frac{1}{2}\)….(1)

Also, \(\mathrm{P}\left(\mathrm{A} / \mathrm{E}_2\right)=\frac{1}{8}\) (given), \(\mathrm{P}\left(\mathrm{A} / \mathrm{E}_3\right)=1\)….(2)

(because his answer is correct, given that he knows it]

CBSE Important Questions for Class 12 Maths

and \(P\left(A / E_1\right)=\frac{1}{4}\)…..(3)

(his answer is correct, given that he guesses and the multiple choice question has four choices)

1. (2)Required probability = P(E₃) = 1/2 (from eq(1))

2. (1)Required probability = P( A/E₃) = 1 (from eq(2))

3. (1) Required probability = P( A/E₁) = 1/4 (from eq(3))

4. (1) Required probability = P(E₃/A)

= \(\frac{P\left(E_3\right) \cdot P\left(A / E_3\right)}{P\left(E_1\right) \cdot P\left(A / E_1\right)+P\left(E_2\right) \cdot P\left(A / E_2\right)+P\left(E_3\right) \cdot P\left(A / E_3\right)}\) (By Baye’s Theorem)

= \(\frac{\left(\frac{1}{2} \times 1\right)}{\left(\frac{1}{3} \times \frac{1}{4}\right)+\left(\frac{1}{6} \times \frac{1}{8}\right)+\left(\frac{1}{2} \times 1\right)}=\frac{\frac{1}{2}}{\left(\frac{1}{12}+\frac{1}{48}+\frac{1}{2}\right)}\)

= \(\frac{\frac{1}{2}}{\left(\frac{4+1+24}{48}\right)}=\frac{1}{2} \times \frac{48}{29}=\frac{24}{29}\)

5. (3) \(\sum_{k=1}^3 \mathrm{P}\left(\mathrm{E}_2 \mid \mathrm{A}\right)\)

= \(\mathrm{P}\left(\mathrm{E}_1 / \mathrm{A}\right)+\mathrm{P}\left(\mathrm{E}_2 / \mathrm{A}\right)+\mathrm{P}\left(\mathrm{E}_2 / \mathrm{A}\right)=\frac{\mathrm{P}\left(\mathrm{E}_1\right) \cdot \mathrm{P}\left(\mathrm{A} / \mathrm{E}_1\right)+\mathrm{P}\left(\mathrm{E}_2\right) \cdot \mathrm{P}\left(\mathrm{A} / \mathrm{E}_2\right)+\mathrm{P}\left(\mathrm{E}_3\right) \cdot \mathrm{P}\left(\mathrm{A} / \mathrm{E}_3\right)}{\mathrm{P}\left(\mathrm{E}_1\right) \cdot \mathrm{P}\left(\mathrm{A} / \mathrm{E}_1\right)+\mathrm{P}\left(\mathrm{E}_2\right) \cdot \mathrm{P}\left(\mathrm{A} / \mathrm{E}_2\right)+\mathrm{P}\left(\mathrm{E}_3\right) \cdot \mathrm{P}\left(\mathrm{A} / \mathrm{E}_3\right)}\)

[By applying Baye’s theorem and then taking L.C.M.] =1

CBSE Important Questions for Class 12 Maths

Question 12. Suppose a girl throws a die. If she gets 1 or 2, she tosses a coin three times and notes the number of tails. If she gets 3, 4. 5 or 6. she tosses a coin once and notes whether a ‘head’ or ‘tail’ is obtained. If she obtained exactly one ‘tail1, what is the probability that she threw 3, 4.. 5 or 6 with the die?

Solution:

When a die is thrown then the sample space contains 6 outcomes i.e. S= {1,2, 4, 5, 6}

Let E₁: Event that she gets 1 or 2 on the die,

E₂: Event that she gets 3, 4, 5 or 6 on the die.

A: Event that she acts exactly one tail

Here, \(\mathrm{P}\left(\mathrm{E}_1\right)=\frac{2}{6}=\frac{1}{3}, P\left(\mathrm{E}_2\right)=\frac{4}{6}=\frac{2}{3} \text { and } \mathrm{P}\left(\mathrm{A} / \mathrm{E}_1\right)=\frac{3}{8}, \mathrm{P}\left(\mathrm{A} / \mathrm{E}_2\right)=\frac{1}{2}\)

When she gets 1 or 2, then she throws a coin three times, and the outcomes are {HHH, TTT, HHT, THH, HTH, TTH, HTT, THT}

⇒ Required probability = P(E₂/A)

= \(\frac{P\left(E_2\right) \cdot P\left(A / E_2\right)}{P\left(E_1\right) P\left(A / E_1\right)+P\left(E_2\right) \cdot P\left(A / E_2\right)} \text { (Using Baye’s Theorem) } \)

= \(\frac{\frac{2}{3} \times \frac{1}{2}}{\left(\frac{1}{3} \times \frac{3}{8}\right)+\left(\frac{2}{3} \times \frac{1}{2}\right)}=\frac{\frac{1}{3}}{\left(\frac{1}{8}+\frac{1}{3}\right)}=\frac{8}{11}\)

Question 13. Two numbers are selected at random (without replacement) from the first six positive integers. Let X denote the larger of the two numbers obtained. Find the probability distribution of the random variable X. and hence find the mean of the distribution. [CBSE 2018|

Solution:

S = {1, 2, 3, 4, 5, 6}

X denotes the larger of the two numbers (given)

If X = 2, then favourable cases are {(1, 2),(2,1)}

If X = 3, then favourable cases are {(1, 3),(2, 3), (3, 1),(3, 2)}

If X = 4, then favourable cases are {(1, 4),(2, 4),(3, 4),(4, 1),(4. 2),(4, 3)}

If X = 5, then favourable cases are {(1, 5),(2, 5),(3, 5),(4, 5),(5,1),(5, 2),(5, 3),(5, 4)}

If X = 6, then favourable cases are {(1,6),(2,6),(3,6),(4,6),(5,6),(6, 1),(6,2),(6,3),(6,4),(6,5)}

⇒ Mean = \(\Sigma X . P(X)=\frac{4+12+24+40+60}{30}=\frac{140}{30}=\frac{14}{3}\)

Question 14. A bag contains 5 red and 4 black balls, a second bag contains 3 red and 6 black balls. One of the two bags is selected at random and two balls are drawn at random (without replacement), both of which are found to be red. Find the probability that the balls are drawn from the second bag.

Solution:

Let E₁: Event that first bag is selected,

E₂: Event that the second bag is selected,

E: Event that both drawn balls are red

∴ Required probability = \(\mathrm{P}\left(\mathrm{E}_2 / \mathrm{E}\right)\)

CBSE Important Questions for Class 12 Maths

= \(\frac{P\left(E_2\right) \cdot P\left(E_{/} / E_2\right)}{P\left(E_1\right) \cdot P\left(E / E_1\right)+P\left(E_2\right) \cdot P\left(E / E_2\right)}\)(Using Bayes’ Theorem)

= \(\frac{\frac{1}{2} \times \frac{1}{12}}{\frac{1}{2} \times \frac{5}{18}+\frac{1}{2} \times \frac{1}{12}}=\left(\frac{\frac{1}{12}}{\frac{5}{18}+\frac{1}{12}}\right)=\frac{3}{13}\)