Let's go through each problem one by one and verify the given solutions:

Problem 1:

Question: An urn contains 5 balls. Two balls are drawn and are found to be white. What is the probability of all the balls being white?

Solution Verification: Let’s define:

• $W$: Event that all balls are white.
• $D$: Event that two balls drawn are white.

The problem asks for $P(W|D)$, which is the probability that all the balls are white given that two drawn balls are white.

Using Bayes' theorem: $P(W|D) = \frac{P(D|W) \cdot P(W)}{P(D)}$

Assuming two cases:

1. All 5 balls are white: $P(W) = \frac{1}{2}$ (equal chances).
2. Not all balls are white: Probability of drawing two white balls is zero.

Since drawing two white balls is observed: $P(D|W) = 1, \quad P(D|\neg W) = 0$

So, $P(W|D) = \frac{1 \cdot \frac{1}{2}}{\frac{1}{2}} = \frac{1}{2}$.

Answer is correct: $\frac{1}{2}$.

Problem 2:

Question: The chances of three candidates A, B, and C to become the manager of a company are in the ratio 3:5:4. The probabilities of introducing a special bonus scheme by them if selected are 0.6, 0.4, and 0.5 respectively. If the bonus scheme is introduced, what is the probability that B has become the manager?

Solution Verification: Define events:

• $A, B, C$: Probability of becoming the manager.
• $S$: Bonus scheme is introduced.

Given ratios: $P(A) = \frac{3}{12}, \quad P(B) = \frac{5}{12}, \quad P(C) = \frac{4}{12}.$

Probabilities of bonus scheme if they become manager: $P(S|A) = 0.6, \quad P(S|B) = 0.4, \quad P(S|C) = 0.5.$

Using the total probability theorem: $P(S) = P(S|A) \cdot P(A) + P(S|B) \cdot P(B) + P(S|C) \cdot P(C).$

Calculating $P(S)$: $P(S) = 0.6 \cdot \frac{3}{12} + 0.4 \cdot \frac{5}{12} + 0.5 \cdot \frac{4}{12} = 0.15 + 0.1667 + 0.1667 = 0.4834.$

Probability that B is manager given the scheme: $P(B|S) = \frac{P(S|B) \cdot P(B)}{P(S)} = \frac{0.4 \cdot \frac{5}{12}}{0.4834} \approx 0.3448.$

Answer is correct: $0.3448$.

Problem 3:

Question: There are three unbiased coins and one biased coin with heads on both sides. A coin is chosen at random and tossed 4 times. If heads occur all 4 times, what is the probability that the biased coin has been chosen?

Solution Verification: Define events:

• $B$: Biased coin chosen.
• $U$: Unbiased coin chosen.
• $H_4$: Four heads in a row.

Probabilities: $P(B) = \frac{1}{4}, \quad P(U) = \frac{3}{4}.$

Probability of 4 heads: $P(H_4|B) = 1, \quad P(H_4|U) = \left(\frac{1}{2}\right)^4 = \frac{1}{16}.$

Using Bayes' theorem: $P(B|H_4) = \frac{P(H_4|B) \cdot P(B)}{P(H_4|B) \cdot P(B) + P(H_4|U) \cdot P(U)} = \frac{1 \cdot \frac{1}{4}}{\frac{1}{4} + \frac{1}{16} \cdot \frac{3}{4}} = \frac{16}{19}.$

Answer is correct: $\frac{16}{19}$.

Would you like more details on any problem? Here are some related questions:

1. What is Bayes' theorem, and how is it used in probability problems?
2. How do you calculate probabilities when given ratios?
3. How can the total probability theorem help in probability calculations?
4. What are unbiased and biased coins, and how do they affect probability?
5. What are conditional probabilities, and why are they important in probability problems?

Tip: Always define all events and conditions clearly when solving probability problems; this helps to avoid confusion and ensures accurate calculations.