Questions & explanations
1. Box1 has 5 white, 6 black balls. Box2 has 5 white, 3 black balls. One ball is transferred from Box1 to Box2, then a ball is drawn from Box2 and found white. What is the probability the transferred ball was white?
- 25/49
- 5/11
- 25/88
- 49/88
Answer: 25/49
Let E1 = transferred ball white, E2 = transferred ball black. Priors: P(E1)=5/11, P(E2)=6/11. Likelihoods: P(white drawn|E1)=5/8, P(white drawn|E2)=4/8=1/2. Total probability P(white)=49/88 from earlier. By Bayes' theorem, P(E1|white)= (5/11)*(5/8) / (49/88) = (25/88)/(49/88)=25/49.
2. A, B, C, D are mutually independent with P(A)=1/2, P(B)=1/3, P(C)=1/4, P(D)=1/5. What is P(exactly one occurs)?
- 77/60
- 4/5
- 1/2
- 5/12
Answer: 5/12
Using independence, P(exactly one) = P(A)P(B')P(C')P(D') + P(A')P(B)P(C')P(D') + P(A')P(B')P(C)P(D') + P(A')P(B')P(C')P(D). Compute: (1/2)(2/3)(3/4)(4/5)=1/5; (1/2)(1/3)(3/4)(4/5)=1/10; (1/2)(2/3)(1/4)(4/5)=1/15; (1/2)(2/3)(3/4)(1/5)=1/20. Sum = 1/5+1/10+1/15+1/20 = 5/12.
3. Bag1 has 4 red, 6 white; Bag2 has 6 red, 4 white. A bag is chosen at random, then two balls are drawn without replacement. Both are red. What is the probability that Bag1 was chosen?
- 1/2
- 2/7
- 1/3
- 3/7
Answer: 2/7
Using Bayes' theorem: P(Bag1|both red) = P(both red|Bag1)P(Bag1) / P(both red). P(both red|Bag1) = (4/10)(3/9)=2/15; P(both red|Bag2) = (6/10)(5/9)=1/3; P(both red) = (1/2)(2/15)+(1/2)(1/3)=7/30. So posterior = (1/2)(2/15)/(7/30)=2/7.
4. Two fair dice are rolled. What is the probability that the sum is 10 given that the first die shows 5?
- 1/3
- 1/6
- 1/12
- 1/36
Answer: 1/6
Given first die is 5, the sample space reduces to 6 outcomes: (5,1) to (5,6). Only (5,5) gives sum 10. So probability = 1/6. Alternatively, using formula: P(sum=10 | first=5) = P(sum=10 and first=5)/P(first=5) = (1/36)/(6/36) = 1/6.
5. A fair coin (P(H)=1/2) and a biased coin (P(H)=3/4) are equally likely. One coin is chosen at random and tossed twice, both times showing heads. What is the probability that the biased coin was chosen?
- 9/25
- 3/5
- 9/13
- 1/2
Answer: 9/13
Using Bayes' theorem sequentially: after first head, P(Coin2|H) = (1/2×3/4)/(1/2×1/2+1/2×3/4) = 3/5. This becomes prior for second toss. Then P(Coin2|HH) = (3/5×3/4)/(3/5×3/4+2/5×1/2) = (9/20)/(9/20+1/5) = (9/20)/(13/20) = 9/13.
6. A disease affects 1% of the population. A test has 99% sensitivity and 95% specificity. A person takes two independent tests and both are positive. What is the probability they have the disease?
- 0.980
- 0.167
- 0.798
- 0.500
Answer: 0.798
Using Bayes' theorem: P(D)=0.01, P(+|D)=0.99, P(+|~D)=0.05. For two independent positives, P(++|D)=0.99^2=0.9801, P(++|~D)=0.05^2=0.0025. Total P(++)=0.01*0.9801+0.99*0.0025=0.012276. Posterior P(D|++)=0.009801/0.012276≈0.798.
7. If A and B are independent events, which of the following is also independent?
- A and B' only
- All of A and B', A' and B, A' and B'
- A' and B' only
- A' and B only
Answer: All of A and B', A' and B, A' and B'
If A and B are independent, then P(A∩B') = P(A) - P(A∩B) = P(A) - P(A)P(B) = P(A)(1-P(B)) = P(A)P(B'), so A and B' are independent. Similarly, A' and B, and A' and B' are independent. Thus all three pairs are independent.
8. In a tree diagram for a two-stage experiment, which of the following is true about the probability of a path?
- It is the sum of the probabilities along the branches.
- It is the difference of the probabilities along the branches.
- It is the product of the probabilities along the branches.
- It is the average of the probabilities along the branches.
Answer: It is the product of the probabilities along the branches.
In a tree diagram, the probability of a path (a sequence of events) is the product of the probabilities along the branches, because each branch represents a conditional probability and the multiplication theorem applies.
9. An urn has 4 red and 6 white balls. Three balls are drawn successively without replacement. What is the probability of getting exactly 2 red balls?
- 1/10
- 3/10
- 1/6
- 2/5
Answer: 3/10
Exactly 2 reds in 3 draws can occur in 3 orders: RRW, RWR, WRR. Each order has probability (4/10)*(3/9)*(6/8)=72/720=1/10. Summing gives 3/10. Alternatively, using combinations: C(4,2)*C(6,1)/C(10,3)=6*6/120=36/120=3/10.
10. A and B solve a problem independently with probabilities 1/2 and 1/3. Given that the problem is solved, what is the probability that both solved it?
- 1/4
- 1/6
- 1/3
- 1/2
Answer: 1/4
Using conditional probability: P(both solved | solved) = P(A∩B) / P(A∪B). Since independent, P(A∩B) = (1/2)(1/3) = 1/6. P(A∪B) = 1 - P(neither) = 1 - (1/2)(2/3) = 2/3. So required probability = (1/6) / (2/3) = 1/4.
11. A fair die is rolled. Given that the outcome is even, what is the probability that it is 4?
- 1/3
- 1/6
- 1/2
- 2/3
Answer: 1/3
When the die is fair, the sample space has 6 outcomes. Knowing the outcome is even reduces the sample space to {2,4,6}. Among these 3 equally likely outcomes, only 1 is 4, so the conditional probability is 1/3.
12. One card is drawn from a standard deck of 52 cards. What is the probability that it is a king given that it is a face card?
- 1/3
- 1/13
- 1/4
- 1/12
Answer: 1/3
Using conditional probability formula P(A|B) = P(A∩B)/P(B). A = king, B = face card. A∩B = king (since every king is a face card). P(A∩B) = 4/52, P(B) = 12/52. So P(A|B) = (4/52)/(12/52) = 4/12 = 1/3.