1. There are three events A, B, C; one and only one of which must occur. The odds are 8 to 3 against A and 5:2 against B. What are the odds against C? 2. A, B and C are bidding for... 1. There are three events A, B, C; one and only one of which must occur. The odds are 8 to 3 against A and 5:2 against B. What are the odds against C? 2. A, B and C are bidding for a contract. It is believed that A has exactly half the chance that B has, B in turn, is 4/5th as likely as C to win the contract. What is the probability for each to win the contract? Read more

Steps to Solve

1. Convert odds against A to probability of A

Odds against A are 8 to 3, meaning for every 8 unfavorable outcomes, there are 3 favorable outcomes. Thus, the probability of A occurring is:

$P(A) = \frac{3}{8+3} = \frac{3}{11}$

2. Convert odds against B to probability of B

Odds against B are 5 to 2, meaning for every 5 unfavorable outcomes, there are 2 favorable outcomes. Thus, the probability of B occurring is:

$P(B) = \frac{2}{5+2} = \frac{2}{7}$

3. Calculate the probability of C

Since only one of the events A, B, or C must occur, the sum of their probabilities is 1:

$P(A) + P(B) + P(C) = 1$

Substitute the values of $P(A)$ and $P(B)$: $\frac{3}{11} + \frac{2}{7} + P(C) = 1$

Find a common denominator for $\frac{3}{11}$ and $\frac{2}{7}$, which is 77: $\frac{21}{77} + \frac{22}{77} + P(C) = 1$

$\frac{43}{77} + P(C) = 1$

$P(C) = 1 - \frac{43}{77} = \frac{77}{77} - \frac{43}{77} = \frac{34}{77}$

4. Convert the probability of C to odds against C

$P(C) = \frac{34}{77}$

This means there are 34 favorable outcomes out of 77 total outcomes. Thus, there are $77 - 34 = 43$ unfavorable outcomes.

The odds against C are the ratio of unfavorable to favorable outcomes: $43:34$

5. Define the probabilities for A, B, and C in terms of a variable

Let $P(C) = x$. We are given $P(B) = \frac{4}{5}x$. Also, $P(A) = \frac{1}{2}P(B) = \frac{1}{2} \cdot \frac{4}{5}x = \frac{2}{5}x$.

6. Solve for $x$ using the fact that the probabilities must sum to 1

Since A, B, and C are the only possibilities, $P(A) + P(B) + P(C) = 1$.

Substituting the expressions in terms of $x$, we have: $\frac{2}{5}x + \frac{4}{5}x + x = 1$ $\frac{2x + 4x + 5x}{5} = 1$ $\frac{11x}{5} = 1$ $11x = 5$ $x = \frac{5}{11}$

7. Calculate the probability for each event

$P(C) = x = \frac{5}{11}$

$P(B) = \frac{4}{5}x = \frac{4}{5} \cdot \frac{5}{11} = \frac{4}{11}$

$P(A) = \frac{2}{5}x = \frac{2}{5} \cdot \frac{5}{11} = \frac{2}{11}$