Important Questions

1

NCERT Ex 13.1 Q1 • 2 Marks2 Marks

Given P(A) = 7/13, P(B) = 9/13, and P(A ∩ B) = 4/13, evaluate P(A|B) and P(A ∪ B).

▶ Show Detailed Solution

Step 1 (Conditional P(A|B)):
P(A|B) = P(A ∩ B)/P(B) = (4/13) / (9/13) = 4/9.

Step 2 (Union P(A ∪ B)):
P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = 7/13 + 9/13 - 4/13 = 12/13.

✔ Final Answer: P(A|B) = 4/9, P(A ∪ B) = 12/13

2

CBSE Board 2024 6-Marker • 5 Marks5 Marks

Bag I contains 3 red and 4 black balls, while Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and is found to be RED. Find the probability that it was drawn from Bag II.

▶ Show Detailed Solution

Step 1 (Define Events & Priors):
E1 = Bag I, E2 = Bag II ⟹ P(E1) = P(E2) = 1/2.
A = Red ball drawn.
P(A|E1) = 3/7, P(A|E2) = 5/11.

Step 2 (Bayes' Theorem):
P(E2|A) = [ (1/2)(5/11) ] / [ (1/2)(3/7) + (1/2)(5/11) ]
= (5/11) / (3/7 + 5/11) = (5/11) / (68/77) = 35/68.

✔ Final Answer: 35/68

3

CBSE Board 2023 6-Marker • 5 Marks5 Marks

A doctor is to visit a patient. Probabilities that he will come by train, bus, scooter, car are 3/10, 1/5, 1/10, 2/5. Probabilities he will be late are 1/4, 1/3, 1/12, 0. If he arrives late, what is prob he came by train?

▶ Show Detailed Solution

Step 1 (Bayes Setup):
P(T) = 3/10, P(B) = 2/10, P(S) = 1/10, P(C) = 4/10.
P(L|T) = 1/4, P(L|B) = 1/3, P(L|S) = 1/12, P(L|C) = 0.

Step 2 (Calculate P(Train|Late)):
P(Train|Late) = (3/10 * 1/4) / (3/40 + 2/30 + 1/120 + 0) = (3/40) / (18/120) = (3/40) / (3/20) = 1/2.

✔ Final Answer: 1/2

4

CBSE Board 2022 • 5 Marks5 Marks

An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. Probabilities of accidents are 0.01, 0.03, 0.15. One insured person meets with accident. Find prob he is scooter driver.

▶ Show Detailed Solution

Step 1 (Priors & Likelihoods):
P(E1) = 2000/12000 = 1/6, P(E2) = 4000/12000 = 1/3, P(E3) = 6000/12000 = 1/2.
P(A|E1) = 0.01, P(A|E2) = 0.03, P(A|E3) = 0.15.

Step 2 (Apply Bayes' Theorem):
P(E1|A) = (1/6 * 0.01) / [ (1/6 * 0.01) + (1/3 * 0.03) + (1/2 * 0.15) ]
= (0.01/6) / [ (0.01 + 0.06 + 0.45)/6 ] = 0.01 / 0.52 = 1/52.

✔ Final Answer: 1/52

5

NCERT Ex 13.1 Q1 (Variant #5) • 2 Marks2 Marks

Given P(A) = 7/13, P(B) = 9/13, and P(A ∩ B) = 4/13, evaluate P(A|B) and P(A ∪ B).

▶ Show Detailed Solution

Step 1 (Conditional P(A|B)):
P(A|B) = P(A ∩ B)/P(B) = (4/13) / (9/13) = 4/9.

Step 2 (Union P(A ∪ B)):
P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = 7/13 + 9/13 - 4/13 = 12/13.

✔ Final Answer: P(A|B) = 4/9, P(A ∪ B) = 12/13

6

CBSE Board 2024 6-Marker (Variant #6) • 5 Marks5 Marks

Bag I contains 3 red and 4 black balls, while Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and is found to be RED. Find the probability that it was drawn from Bag II.

▶ Show Detailed Solution

Step 1 (Define Events & Priors):
E1 = Bag I, E2 = Bag II ⟹ P(E1) = P(E2) = 1/2.
A = Red ball drawn.
P(A|E1) = 3/7, P(A|E2) = 5/11.

Step 2 (Bayes' Theorem):
P(E2|A) = [ (1/2)(5/11) ] / [ (1/2)(3/7) + (1/2)(5/11) ]
= (5/11) / (3/7 + 5/11) = (5/11) / (68/77) = 35/68.

✔ Final Answer: 35/68

7

CBSE Board 2023 6-Marker (Variant #7) • 5 Marks5 Marks

A doctor is to visit a patient. Probabilities that he will come by train, bus, scooter, car are 3/10, 1/5, 1/10, 2/5. Probabilities he will be late are 1/4, 1/3, 1/12, 0. If he arrives late, what is prob he came by train?

▶ Show Detailed Solution

Step 1 (Bayes Setup):
P(T) = 3/10, P(B) = 2/10, P(S) = 1/10, P(C) = 4/10.
P(L|T) = 1/4, P(L|B) = 1/3, P(L|S) = 1/12, P(L|C) = 0.

Step 2 (Calculate P(Train|Late)):
P(Train|Late) = (3/10 * 1/4) / (3/40 + 2/30 + 1/120 + 0) = (3/40) / (18/120) = (3/40) / (3/20) = 1/2.

✔ Final Answer: 1/2

8

CBSE Board 2022 (Variant #8) • 5 Marks5 Marks

An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. Probabilities of accidents are 0.01, 0.03, 0.15. One insured person meets with accident. Find prob he is scooter driver.

▶ Show Detailed Solution

Step 1 (Priors & Likelihoods):
P(E1) = 2000/12000 = 1/6, P(E2) = 4000/12000 = 1/3, P(E3) = 6000/12000 = 1/2.
P(A|E1) = 0.01, P(A|E2) = 0.03, P(A|E3) = 0.15.

Step 2 (Apply Bayes' Theorem):
P(E1|A) = (1/6 * 0.01) / [ (1/6 * 0.01) + (1/3 * 0.03) + (1/2 * 0.15) ]
= (0.01/6) / [ (0.01 + 0.06 + 0.45)/6 ] = 0.01 / 0.52 = 1/52.

✔ Final Answer: 1/52

9

NCERT Ex 13.1 Q1 (Variant #9) • 2 Marks2 Marks

Given P(A) = 7/13, P(B) = 9/13, and P(A ∩ B) = 4/13, evaluate P(A|B) and P(A ∪ B).

▶ Show Detailed Solution

Step 1 (Conditional P(A|B)):
P(A|B) = P(A ∩ B)/P(B) = (4/13) / (9/13) = 4/9.

Step 2 (Union P(A ∪ B)):
P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = 7/13 + 9/13 - 4/13 = 12/13.

✔ Final Answer: P(A|B) = 4/9, P(A ∪ B) = 12/13

10

CBSE Board 2024 6-Marker (Variant #10) • 5 Marks5 Marks

Bag I contains 3 red and 4 black balls, while Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and is found to be RED. Find the probability that it was drawn from Bag II.

▶ Show Detailed Solution

Step 1 (Define Events & Priors):
E1 = Bag I, E2 = Bag II ⟹ P(E1) = P(E2) = 1/2.
A = Red ball drawn.
P(A|E1) = 3/7, P(A|E2) = 5/11.

Step 2 (Bayes' Theorem):
P(E2|A) = [ (1/2)(5/11) ] / [ (1/2)(3/7) + (1/2)(5/11) ]
= (5/11) / (3/7 + 5/11) = (5/11) / (68/77) = 35/68.

✔ Final Answer: 35/68

11

CBSE Board 2023 6-Marker (Variant #11) • 5 Marks5 Marks

A doctor is to visit a patient. Probabilities that he will come by train, bus, scooter, car are 3/10, 1/5, 1/10, 2/5. Probabilities he will be late are 1/4, 1/3, 1/12, 0. If he arrives late, what is prob he came by train?

▶ Show Detailed Solution

Step 1 (Bayes Setup):
P(T) = 3/10, P(B) = 2/10, P(S) = 1/10, P(C) = 4/10.
P(L|T) = 1/4, P(L|B) = 1/3, P(L|S) = 1/12, P(L|C) = 0.

Step 2 (Calculate P(Train|Late)):
P(Train|Late) = (3/10 * 1/4) / (3/40 + 2/30 + 1/120 + 0) = (3/40) / (18/120) = (3/40) / (3/20) = 1/2.

✔ Final Answer: 1/2

12

CBSE Board 2022 (Variant #12) • 5 Marks5 Marks

An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. Probabilities of accidents are 0.01, 0.03, 0.15. One insured person meets with accident. Find prob he is scooter driver.

▶ Show Detailed Solution

Step 1 (Priors & Likelihoods):
P(E1) = 2000/12000 = 1/6, P(E2) = 4000/12000 = 1/3, P(E3) = 6000/12000 = 1/2.
P(A|E1) = 0.01, P(A|E2) = 0.03, P(A|E3) = 0.15.

Step 2 (Apply Bayes' Theorem):
P(E1|A) = (1/6 * 0.01) / [ (1/6 * 0.01) + (1/3 * 0.03) + (1/2 * 0.15) ]
= (0.01/6) / [ (0.01 + 0.06 + 0.45)/6 ] = 0.01 / 0.52 = 1/52.

✔ Final Answer: 1/52

13

NCERT Ex 13.1 Q1 (Variant #13) • 2 Marks2 Marks

Given P(A) = 7/13, P(B) = 9/13, and P(A ∩ B) = 4/13, evaluate P(A|B) and P(A ∪ B).

▶ Show Detailed Solution

Step 1 (Conditional P(A|B)):
P(A|B) = P(A ∩ B)/P(B) = (4/13) / (9/13) = 4/9.

Step 2 (Union P(A ∪ B)):
P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = 7/13 + 9/13 - 4/13 = 12/13.

✔ Final Answer: P(A|B) = 4/9, P(A ∪ B) = 12/13

14

CBSE Board 2024 6-Marker (Variant #14) • 5 Marks5 Marks

Bag I contains 3 red and 4 black balls, while Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and is found to be RED. Find the probability that it was drawn from Bag II.

▶ Show Detailed Solution

Step 1 (Define Events & Priors):
E1 = Bag I, E2 = Bag II ⟹ P(E1) = P(E2) = 1/2.
A = Red ball drawn.
P(A|E1) = 3/7, P(A|E2) = 5/11.

Step 2 (Bayes' Theorem):
P(E2|A) = [ (1/2)(5/11) ] / [ (1/2)(3/7) + (1/2)(5/11) ]
= (5/11) / (3/7 + 5/11) = (5/11) / (68/77) = 35/68.

✔ Final Answer: 35/68

15

CBSE Board 2023 6-Marker (Variant #15) • 5 Marks5 Marks

A doctor is to visit a patient. Probabilities that he will come by train, bus, scooter, car are 3/10, 1/5, 1/10, 2/5. Probabilities he will be late are 1/4, 1/3, 1/12, 0. If he arrives late, what is prob he came by train?

▶ Show Detailed Solution

Step 1 (Bayes Setup):
P(T) = 3/10, P(B) = 2/10, P(S) = 1/10, P(C) = 4/10.
P(L|T) = 1/4, P(L|B) = 1/3, P(L|S) = 1/12, P(L|C) = 0.

Step 2 (Calculate P(Train|Late)):
P(Train|Late) = (3/10 * 1/4) / (3/40 + 2/30 + 1/120 + 0) = (3/40) / (18/120) = (3/40) / (3/20) = 1/2.

✔ Final Answer: 1/2

16

CBSE Board 2022 (Variant #16) • 5 Marks5 Marks

An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. Probabilities of accidents are 0.01, 0.03, 0.15. One insured person meets with accident. Find prob he is scooter driver.

▶ Show Detailed Solution

Step 1 (Priors & Likelihoods):
P(E1) = 2000/12000 = 1/6, P(E2) = 4000/12000 = 1/3, P(E3) = 6000/12000 = 1/2.
P(A|E1) = 0.01, P(A|E2) = 0.03, P(A|E3) = 0.15.

Step 2 (Apply Bayes' Theorem):
P(E1|A) = (1/6 * 0.01) / [ (1/6 * 0.01) + (1/3 * 0.03) + (1/2 * 0.15) ]
= (0.01/6) / [ (0.01 + 0.06 + 0.45)/6 ] = 0.01 / 0.52 = 1/52.

✔ Final Answer: 1/52

17

NCERT Ex 13.1 Q1 (Variant #17) • 2 Marks2 Marks

Given P(A) = 7/13, P(B) = 9/13, and P(A ∩ B) = 4/13, evaluate P(A|B) and P(A ∪ B).

▶ Show Detailed Solution

Step 1 (Conditional P(A|B)):
P(A|B) = P(A ∩ B)/P(B) = (4/13) / (9/13) = 4/9.

Step 2 (Union P(A ∪ B)):
P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = 7/13 + 9/13 - 4/13 = 12/13.

✔ Final Answer: P(A|B) = 4/9, P(A ∪ B) = 12/13

18

CBSE Board 2024 6-Marker (Variant #18) • 5 Marks5 Marks

Bag I contains 3 red and 4 black balls, while Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and is found to be RED. Find the probability that it was drawn from Bag II.

▶ Show Detailed Solution

Step 1 (Define Events & Priors):
E1 = Bag I, E2 = Bag II ⟹ P(E1) = P(E2) = 1/2.
A = Red ball drawn.
P(A|E1) = 3/7, P(A|E2) = 5/11.

Step 2 (Bayes' Theorem):
P(E2|A) = [ (1/2)(5/11) ] / [ (1/2)(3/7) + (1/2)(5/11) ]
= (5/11) / (3/7 + 5/11) = (5/11) / (68/77) = 35/68.

✔ Final Answer: 35/68

19

CBSE Board 2023 6-Marker (Variant #19) • 5 Marks5 Marks

A doctor is to visit a patient. Probabilities that he will come by train, bus, scooter, car are 3/10, 1/5, 1/10, 2/5. Probabilities he will be late are 1/4, 1/3, 1/12, 0. If he arrives late, what is prob he came by train?

▶ Show Detailed Solution

Step 1 (Bayes Setup):
P(T) = 3/10, P(B) = 2/10, P(S) = 1/10, P(C) = 4/10.
P(L|T) = 1/4, P(L|B) = 1/3, P(L|S) = 1/12, P(L|C) = 0.

Step 2 (Calculate P(Train|Late)):
P(Train|Late) = (3/10 * 1/4) / (3/40 + 2/30 + 1/120 + 0) = (3/40) / (18/120) = (3/40) / (3/20) = 1/2.

✔ Final Answer: 1/2

20

CBSE Board 2022 (Variant #20) • 5 Marks5 Marks

An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. Probabilities of accidents are 0.01, 0.03, 0.15. One insured person meets with accident. Find prob he is scooter driver.

▶ Show Detailed Solution

Step 1 (Priors & Likelihoods):
P(E1) = 2000/12000 = 1/6, P(E2) = 4000/12000 = 1/3, P(E3) = 6000/12000 = 1/2.
P(A|E1) = 0.01, P(A|E2) = 0.03, P(A|E3) = 0.15.

Step 2 (Apply Bayes' Theorem):
P(E1|A) = (1/6 * 0.01) / [ (1/6 * 0.01) + (1/3 * 0.03) + (1/2 * 0.15) ]
= (0.01/6) / [ (0.01 + 0.06 + 0.45)/6 ] = 0.01 / 0.52 = 1/52.

✔ Final Answer: 1/52

21

NCERT Ex 13.1 Q1 (Variant #21) • 2 Marks2 Marks

Given P(A) = 7/13, P(B) = 9/13, and P(A ∩ B) = 4/13, evaluate P(A|B) and P(A ∪ B).

▶ Show Detailed Solution

Step 1 (Conditional P(A|B)):
P(A|B) = P(A ∩ B)/P(B) = (4/13) / (9/13) = 4/9.

Step 2 (Union P(A ∪ B)):
P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = 7/13 + 9/13 - 4/13 = 12/13.

✔ Final Answer: P(A|B) = 4/9, P(A ∪ B) = 12/13

22

CBSE Board 2024 6-Marker (Variant #22) • 5 Marks5 Marks

Bag I contains 3 red and 4 black balls, while Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and is found to be RED. Find the probability that it was drawn from Bag II.

▶ Show Detailed Solution

Step 1 (Define Events & Priors):
E1 = Bag I, E2 = Bag II ⟹ P(E1) = P(E2) = 1/2.
A = Red ball drawn.
P(A|E1) = 3/7, P(A|E2) = 5/11.

Step 2 (Bayes' Theorem):
P(E2|A) = [ (1/2)(5/11) ] / [ (1/2)(3/7) + (1/2)(5/11) ]
= (5/11) / (3/7 + 5/11) = (5/11) / (68/77) = 35/68.

✔ Final Answer: 35/68

23

CBSE Board 2023 6-Marker (Variant #23) • 5 Marks5 Marks

A doctor is to visit a patient. Probabilities that he will come by train, bus, scooter, car are 3/10, 1/5, 1/10, 2/5. Probabilities he will be late are 1/4, 1/3, 1/12, 0. If he arrives late, what is prob he came by train?

▶ Show Detailed Solution

Step 1 (Bayes Setup):
P(T) = 3/10, P(B) = 2/10, P(S) = 1/10, P(C) = 4/10.
P(L|T) = 1/4, P(L|B) = 1/3, P(L|S) = 1/12, P(L|C) = 0.

Step 2 (Calculate P(Train|Late)):
P(Train|Late) = (3/10 * 1/4) / (3/40 + 2/30 + 1/120 + 0) = (3/40) / (18/120) = (3/40) / (3/20) = 1/2.

✔ Final Answer: 1/2

24

CBSE Board 2022 (Variant #24) • 5 Marks5 Marks

An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. Probabilities of accidents are 0.01, 0.03, 0.15. One insured person meets with accident. Find prob he is scooter driver.

▶ Show Detailed Solution

Step 1 (Priors & Likelihoods):
P(E1) = 2000/12000 = 1/6, P(E2) = 4000/12000 = 1/3, P(E3) = 6000/12000 = 1/2.
P(A|E1) = 0.01, P(A|E2) = 0.03, P(A|E3) = 0.15.

Step 2 (Apply Bayes' Theorem):
P(E1|A) = (1/6 * 0.01) / [ (1/6 * 0.01) + (1/3 * 0.03) + (1/2 * 0.15) ]
= (0.01/6) / [ (0.01 + 0.06 + 0.45)/6 ] = 0.01 / 0.52 = 1/52.

✔ Final Answer: 1/52

25

NCERT Ex 13.1 Q1 (Variant #25) • 2 Marks2 Marks

Given P(A) = 7/13, P(B) = 9/13, and P(A ∩ B) = 4/13, evaluate P(A|B) and P(A ∪ B).

▶ Show Detailed Solution

Step 1 (Conditional P(A|B)):
P(A|B) = P(A ∩ B)/P(B) = (4/13) / (9/13) = 4/9.

Step 2 (Union P(A ∪ B)):
P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = 7/13 + 9/13 - 4/13 = 12/13.

✔ Final Answer: P(A|B) = 4/9, P(A ∪ B) = 12/13

26

CBSE Board 2024 6-Marker (Variant #26) • 5 Marks5 Marks

Bag I contains 3 red and 4 black balls, while Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and is found to be RED. Find the probability that it was drawn from Bag II.

▶ Show Detailed Solution

Step 1 (Define Events & Priors):
E1 = Bag I, E2 = Bag II ⟹ P(E1) = P(E2) = 1/2.
A = Red ball drawn.
P(A|E1) = 3/7, P(A|E2) = 5/11.

Step 2 (Bayes' Theorem):
P(E2|A) = [ (1/2)(5/11) ] / [ (1/2)(3/7) + (1/2)(5/11) ]
= (5/11) / (3/7 + 5/11) = (5/11) / (68/77) = 35/68.

✔ Final Answer: 35/68

27

CBSE Board 2023 6-Marker (Variant #27) • 5 Marks5 Marks

A doctor is to visit a patient. Probabilities that he will come by train, bus, scooter, car are 3/10, 1/5, 1/10, 2/5. Probabilities he will be late are 1/4, 1/3, 1/12, 0. If he arrives late, what is prob he came by train?

▶ Show Detailed Solution

Step 1 (Bayes Setup):
P(T) = 3/10, P(B) = 2/10, P(S) = 1/10, P(C) = 4/10.
P(L|T) = 1/4, P(L|B) = 1/3, P(L|S) = 1/12, P(L|C) = 0.

Step 2 (Calculate P(Train|Late)):
P(Train|Late) = (3/10 * 1/4) / (3/40 + 2/30 + 1/120 + 0) = (3/40) / (18/120) = (3/40) / (3/20) = 1/2.

✔ Final Answer: 1/2

28

CBSE Board 2022 (Variant #28) • 5 Marks5 Marks

An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. Probabilities of accidents are 0.01, 0.03, 0.15. One insured person meets with accident. Find prob he is scooter driver.

▶ Show Detailed Solution

Step 1 (Priors & Likelihoods):
P(E1) = 2000/12000 = 1/6, P(E2) = 4000/12000 = 1/3, P(E3) = 6000/12000 = 1/2.
P(A|E1) = 0.01, P(A|E2) = 0.03, P(A|E3) = 0.15.

Step 2 (Apply Bayes' Theorem):
P(E1|A) = (1/6 * 0.01) / [ (1/6 * 0.01) + (1/3 * 0.03) + (1/2 * 0.15) ]
= (0.01/6) / [ (0.01 + 0.06 + 0.45)/6 ] = 0.01 / 0.52 = 1/52.

✔ Final Answer: 1/52

29

NCERT Ex 13.1 Q1 (Variant #29) • 2 Marks2 Marks

Given P(A) = 7/13, P(B) = 9/13, and P(A ∩ B) = 4/13, evaluate P(A|B) and P(A ∪ B).

▶ Show Detailed Solution

Step 1 (Conditional P(A|B)):
P(A|B) = P(A ∩ B)/P(B) = (4/13) / (9/13) = 4/9.

Step 2 (Union P(A ∪ B)):
P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = 7/13 + 9/13 - 4/13 = 12/13.

✔ Final Answer: P(A|B) = 4/9, P(A ∪ B) = 12/13

30

CBSE Board 2024 6-Marker (Variant #30) • 5 Marks5 Marks

Bag I contains 3 red and 4 black balls, while Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and is found to be RED. Find the probability that it was drawn from Bag II.

▶ Show Detailed Solution

Step 1 (Define Events & Priors):
E1 = Bag I, E2 = Bag II ⟹ P(E1) = P(E2) = 1/2.
A = Red ball drawn.
P(A|E1) = 3/7, P(A|E2) = 5/11.

Step 2 (Bayes' Theorem):
P(E2|A) = [ (1/2)(5/11) ] / [ (1/2)(3/7) + (1/2)(5/11) ]
= (5/11) / (3/7 + 5/11) = (5/11) / (68/77) = 35/68.

✔ Final Answer: 35/68

31

CBSE Board 2023 6-Marker (Variant #31) • 5 Marks5 Marks

A doctor is to visit a patient. Probabilities that he will come by train, bus, scooter, car are 3/10, 1/5, 1/10, 2/5. Probabilities he will be late are 1/4, 1/3, 1/12, 0. If he arrives late, what is prob he came by train?

▶ Show Detailed Solution

Step 1 (Bayes Setup):
P(T) = 3/10, P(B) = 2/10, P(S) = 1/10, P(C) = 4/10.
P(L|T) = 1/4, P(L|B) = 1/3, P(L|S) = 1/12, P(L|C) = 0.

Step 2 (Calculate P(Train|Late)):
P(Train|Late) = (3/10 * 1/4) / (3/40 + 2/30 + 1/120 + 0) = (3/40) / (18/120) = (3/40) / (3/20) = 1/2.

✔ Final Answer: 1/2

32

CBSE Board 2022 (Variant #32) • 5 Marks5 Marks

An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. Probabilities of accidents are 0.01, 0.03, 0.15. One insured person meets with accident. Find prob he is scooter driver.

▶ Show Detailed Solution

Step 1 (Priors & Likelihoods):
P(E1) = 2000/12000 = 1/6, P(E2) = 4000/12000 = 1/3, P(E3) = 6000/12000 = 1/2.
P(A|E1) = 0.01, P(A|E2) = 0.03, P(A|E3) = 0.15.

Step 2 (Apply Bayes' Theorem):
P(E1|A) = (1/6 * 0.01) / [ (1/6 * 0.01) + (1/3 * 0.03) + (1/2 * 0.15) ]
= (0.01/6) / [ (0.01 + 0.06 + 0.45)/6 ] = 0.01 / 0.52 = 1/52.

✔ Final Answer: 1/52

33

NCERT Ex 13.1 Q1 (Variant #33) • 2 Marks2 Marks

Given P(A) = 7/13, P(B) = 9/13, and P(A ∩ B) = 4/13, evaluate P(A|B) and P(A ∪ B).

▶ Show Detailed Solution

Step 1 (Conditional P(A|B)):
P(A|B) = P(A ∩ B)/P(B) = (4/13) / (9/13) = 4/9.

Step 2 (Union P(A ∪ B)):
P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = 7/13 + 9/13 - 4/13 = 12/13.

✔ Final Answer: P(A|B) = 4/9, P(A ∪ B) = 12/13

34

CBSE Board 2024 6-Marker (Variant #34) • 5 Marks5 Marks

Bag I contains 3 red and 4 black balls, while Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and is found to be RED. Find the probability that it was drawn from Bag II.

▶ Show Detailed Solution

Step 1 (Define Events & Priors):
E1 = Bag I, E2 = Bag II ⟹ P(E1) = P(E2) = 1/2.
A = Red ball drawn.
P(A|E1) = 3/7, P(A|E2) = 5/11.

Step 2 (Bayes' Theorem):
P(E2|A) = [ (1/2)(5/11) ] / [ (1/2)(3/7) + (1/2)(5/11) ]
= (5/11) / (3/7 + 5/11) = (5/11) / (68/77) = 35/68.

✔ Final Answer: 35/68

35

CBSE Board 2023 6-Marker (Variant #35) • 5 Marks5 Marks

A doctor is to visit a patient. Probabilities that he will come by train, bus, scooter, car are 3/10, 1/5, 1/10, 2/5. Probabilities he will be late are 1/4, 1/3, 1/12, 0. If he arrives late, what is prob he came by train?

▶ Show Detailed Solution

Step 1 (Bayes Setup):
P(T) = 3/10, P(B) = 2/10, P(S) = 1/10, P(C) = 4/10.
P(L|T) = 1/4, P(L|B) = 1/3, P(L|S) = 1/12, P(L|C) = 0.

Step 2 (Calculate P(Train|Late)):
P(Train|Late) = (3/10 * 1/4) / (3/40 + 2/30 + 1/120 + 0) = (3/40) / (18/120) = (3/40) / (3/20) = 1/2.

✔ Final Answer: 1/2

36

CBSE Board 2022 (Variant #36) • 5 Marks5 Marks

An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. Probabilities of accidents are 0.01, 0.03, 0.15. One insured person meets with accident. Find prob he is scooter driver.

▶ Show Detailed Solution

Step 1 (Priors & Likelihoods):
P(E1) = 2000/12000 = 1/6, P(E2) = 4000/12000 = 1/3, P(E3) = 6000/12000 = 1/2.
P(A|E1) = 0.01, P(A|E2) = 0.03, P(A|E3) = 0.15.

Step 2 (Apply Bayes' Theorem):
P(E1|A) = (1/6 * 0.01) / [ (1/6 * 0.01) + (1/3 * 0.03) + (1/2 * 0.15) ]
= (0.01/6) / [ (0.01 + 0.06 + 0.45)/6 ] = 0.01 / 0.52 = 1/52.

✔ Final Answer: 1/52

37

NCERT Ex 13.1 Q1 (Variant #37) • 2 Marks2 Marks

Given P(A) = 7/13, P(B) = 9/13, and P(A ∩ B) = 4/13, evaluate P(A|B) and P(A ∪ B).

▶ Show Detailed Solution

Step 1 (Conditional P(A|B)):
P(A|B) = P(A ∩ B)/P(B) = (4/13) / (9/13) = 4/9.

Step 2 (Union P(A ∪ B)):
P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = 7/13 + 9/13 - 4/13 = 12/13.

✔ Final Answer: P(A|B) = 4/9, P(A ∪ B) = 12/13

38

CBSE Board 2024 6-Marker (Variant #38) • 5 Marks5 Marks

Bag I contains 3 red and 4 black balls, while Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and is found to be RED. Find the probability that it was drawn from Bag II.

▶ Show Detailed Solution

Step 1 (Define Events & Priors):
E1 = Bag I, E2 = Bag II ⟹ P(E1) = P(E2) = 1/2.
A = Red ball drawn.
P(A|E1) = 3/7, P(A|E2) = 5/11.

Step 2 (Bayes' Theorem):
P(E2|A) = [ (1/2)(5/11) ] / [ (1/2)(3/7) + (1/2)(5/11) ]
= (5/11) / (3/7 + 5/11) = (5/11) / (68/77) = 35/68.

✔ Final Answer: 35/68

39

CBSE Board 2023 6-Marker (Variant #39) • 5 Marks5 Marks

A doctor is to visit a patient. Probabilities that he will come by train, bus, scooter, car are 3/10, 1/5, 1/10, 2/5. Probabilities he will be late are 1/4, 1/3, 1/12, 0. If he arrives late, what is prob he came by train?

▶ Show Detailed Solution

Step 1 (Bayes Setup):
P(T) = 3/10, P(B) = 2/10, P(S) = 1/10, P(C) = 4/10.
P(L|T) = 1/4, P(L|B) = 1/3, P(L|S) = 1/12, P(L|C) = 0.

Step 2 (Calculate P(Train|Late)):
P(Train|Late) = (3/10 * 1/4) / (3/40 + 2/30 + 1/120 + 0) = (3/40) / (18/120) = (3/40) / (3/20) = 1/2.

✔ Final Answer: 1/2

40

CBSE Board 2022 (Variant #40) • 5 Marks5 Marks

An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. Probabilities of accidents are 0.01, 0.03, 0.15. One insured person meets with accident. Find prob he is scooter driver.

▶ Show Detailed Solution

Step 1 (Priors & Likelihoods):
P(E1) = 2000/12000 = 1/6, P(E2) = 4000/12000 = 1/3, P(E3) = 6000/12000 = 1/2.
P(A|E1) = 0.01, P(A|E2) = 0.03, P(A|E3) = 0.15.

Step 2 (Apply Bayes' Theorem):
P(E1|A) = (1/6 * 0.01) / [ (1/6 * 0.01) + (1/3 * 0.03) + (1/2 * 0.15) ]
= (0.01/6) / [ (0.01 + 0.06 + 0.45)/6 ] = 0.01 / 0.52 = 1/52.

✔ Final Answer: 1/52

41

NCERT Ex 13.1 Q1 (Variant #41) • 2 Marks2 Marks

Given P(A) = 7/13, P(B) = 9/13, and P(A ∩ B) = 4/13, evaluate P(A|B) and P(A ∪ B).

▶ Show Detailed Solution

Step 1 (Conditional P(A|B)):
P(A|B) = P(A ∩ B)/P(B) = (4/13) / (9/13) = 4/9.

Step 2 (Union P(A ∪ B)):
P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = 7/13 + 9/13 - 4/13 = 12/13.

✔ Final Answer: P(A|B) = 4/9, P(A ∪ B) = 12/13

42

CBSE Board 2024 6-Marker (Variant #42) • 5 Marks5 Marks

Bag I contains 3 red and 4 black balls, while Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and is found to be RED. Find the probability that it was drawn from Bag II.

▶ Show Detailed Solution

Step 1 (Define Events & Priors):
E1 = Bag I, E2 = Bag II ⟹ P(E1) = P(E2) = 1/2.
A = Red ball drawn.
P(A|E1) = 3/7, P(A|E2) = 5/11.

Step 2 (Bayes' Theorem):
P(E2|A) = [ (1/2)(5/11) ] / [ (1/2)(3/7) + (1/2)(5/11) ]
= (5/11) / (3/7 + 5/11) = (5/11) / (68/77) = 35/68.

✔ Final Answer: 35/68

43

CBSE Board 2023 6-Marker (Variant #43) • 5 Marks5 Marks

A doctor is to visit a patient. Probabilities that he will come by train, bus, scooter, car are 3/10, 1/5, 1/10, 2/5. Probabilities he will be late are 1/4, 1/3, 1/12, 0. If he arrives late, what is prob he came by train?

▶ Show Detailed Solution

Step 1 (Bayes Setup):
P(T) = 3/10, P(B) = 2/10, P(S) = 1/10, P(C) = 4/10.
P(L|T) = 1/4, P(L|B) = 1/3, P(L|S) = 1/12, P(L|C) = 0.

Step 2 (Calculate P(Train|Late)):
P(Train|Late) = (3/10 * 1/4) / (3/40 + 2/30 + 1/120 + 0) = (3/40) / (18/120) = (3/40) / (3/20) = 1/2.

✔ Final Answer: 1/2

44

CBSE Board 2022 (Variant #44) • 5 Marks5 Marks

An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. Probabilities of accidents are 0.01, 0.03, 0.15. One insured person meets with accident. Find prob he is scooter driver.

▶ Show Detailed Solution

Step 1 (Priors & Likelihoods):
P(E1) = 2000/12000 = 1/6, P(E2) = 4000/12000 = 1/3, P(E3) = 6000/12000 = 1/2.
P(A|E1) = 0.01, P(A|E2) = 0.03, P(A|E3) = 0.15.

Step 2 (Apply Bayes' Theorem):
P(E1|A) = (1/6 * 0.01) / [ (1/6 * 0.01) + (1/3 * 0.03) + (1/2 * 0.15) ]
= (0.01/6) / [ (0.01 + 0.06 + 0.45)/6 ] = 0.01 / 0.52 = 1/52.

✔ Final Answer: 1/52

45

NCERT Ex 13.1 Q1 (Variant #45) • 2 Marks2 Marks

Given P(A) = 7/13, P(B) = 9/13, and P(A ∩ B) = 4/13, evaluate P(A|B) and P(A ∪ B).

▶ Show Detailed Solution

Step 1 (Conditional P(A|B)):
P(A|B) = P(A ∩ B)/P(B) = (4/13) / (9/13) = 4/9.

Step 2 (Union P(A ∪ B)):
P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = 7/13 + 9/13 - 4/13 = 12/13.

✔ Final Answer: P(A|B) = 4/9, P(A ∪ B) = 12/13

46

CBSE Board 2024 6-Marker (Variant #46) • 5 Marks5 Marks

Bag I contains 3 red and 4 black balls, while Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and is found to be RED. Find the probability that it was drawn from Bag II.

▶ Show Detailed Solution

Step 1 (Define Events & Priors):
E1 = Bag I, E2 = Bag II ⟹ P(E1) = P(E2) = 1/2.
A = Red ball drawn.
P(A|E1) = 3/7, P(A|E2) = 5/11.

Step 2 (Bayes' Theorem):
P(E2|A) = [ (1/2)(5/11) ] / [ (1/2)(3/7) + (1/2)(5/11) ]
= (5/11) / (3/7 + 5/11) = (5/11) / (68/77) = 35/68.

✔ Final Answer: 35/68

47

CBSE Board 2023 6-Marker (Variant #47) • 5 Marks5 Marks

A doctor is to visit a patient. Probabilities that he will come by train, bus, scooter, car are 3/10, 1/5, 1/10, 2/5. Probabilities he will be late are 1/4, 1/3, 1/12, 0. If he arrives late, what is prob he came by train?

▶ Show Detailed Solution

Step 1 (Bayes Setup):
P(T) = 3/10, P(B) = 2/10, P(S) = 1/10, P(C) = 4/10.
P(L|T) = 1/4, P(L|B) = 1/3, P(L|S) = 1/12, P(L|C) = 0.

Step 2 (Calculate P(Train|Late)):
P(Train|Late) = (3/10 * 1/4) / (3/40 + 2/30 + 1/120 + 0) = (3/40) / (18/120) = (3/40) / (3/20) = 1/2.

✔ Final Answer: 1/2

48

CBSE Board 2022 (Variant #48) • 5 Marks5 Marks

An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. Probabilities of accidents are 0.01, 0.03, 0.15. One insured person meets with accident. Find prob he is scooter driver.

▶ Show Detailed Solution

Step 1 (Priors & Likelihoods):
P(E1) = 2000/12000 = 1/6, P(E2) = 4000/12000 = 1/3, P(E3) = 6000/12000 = 1/2.
P(A|E1) = 0.01, P(A|E2) = 0.03, P(A|E3) = 0.15.

Step 2 (Apply Bayes' Theorem):
P(E1|A) = (1/6 * 0.01) / [ (1/6 * 0.01) + (1/3 * 0.03) + (1/2 * 0.15) ]
= (0.01/6) / [ (0.01 + 0.06 + 0.45)/6 ] = 0.01 / 0.52 = 1/52.

✔ Final Answer: 1/52

49

NCERT Ex 13.1 Q1 (Variant #49) • 2 Marks2 Marks

Given P(A) = 7/13, P(B) = 9/13, and P(A ∩ B) = 4/13, evaluate P(A|B) and P(A ∪ B).

▶ Show Detailed Solution

Step 1 (Conditional P(A|B)):
P(A|B) = P(A ∩ B)/P(B) = (4/13) / (9/13) = 4/9.

Step 2 (Union P(A ∪ B)):
P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = 7/13 + 9/13 - 4/13 = 12/13.

✔ Final Answer: P(A|B) = 4/9, P(A ∪ B) = 12/13

50

CBSE Board 2024 6-Marker (Variant #50) • 5 Marks5 Marks

Bag I contains 3 red and 4 black balls, while Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and is found to be RED. Find the probability that it was drawn from Bag II.

▶ Show Detailed Solution

Step 1 (Define Events & Priors):
E1 = Bag I, E2 = Bag II ⟹ P(E1) = P(E2) = 1/2.
A = Red ball drawn.
P(A|E1) = 3/7, P(A|E2) = 5/11.

Step 2 (Bayes' Theorem):
P(E2|A) = [ (1/2)(5/11) ] / [ (1/2)(3/7) + (1/2)(5/11) ]
= (5/11) / (3/7 + 5/11) = (5/11) / (68/77) = 35/68.

✔ Final Answer: 35/68

Probability - Important Questions