Bhavani is going to play a game of chess against one of four opponents in an inter-college sports competition. Each opponent is equally likely to be paired against her. The table below shows the chances of Bhavani losing, when paired against each opponent.
|
Opponent |
Bhavani's Chances of Losing |
|
Opponent 1 |
12% |
|
Opponent 2 |
60% |
|
Opponent 3 |
$x$% |
|
Opponent 4 |
84% |
If the probability that Bhavani Loses the game that day is $\frac{1}{2}$, find the probability for Bhavani to be losing when paired against opponent 3.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $44\%$ ##
Here,
Given, $P(A) = \frac{1}{2}$ (where $A$ is the event of losing the game).
Also, $P(E_1) = P(E_2) = P(E_3) = P(E_4) = \frac{1}{4}$ (probability of being paired with an opponent).
Conditional probabilities of losing:
- $P(A \mid E_1) = 12\% = \frac{12}{100}$
- $P(A \mid E_2) = 60\% = \frac{60}{100}$
- $P(A \mid E_3) = x\% = \frac{x}{100}$
- $P(A \mid E_4) = 84\% = \frac{84}{100}$
Using total probability theorem, we have:
$P(A) = P(E_1) \cdot P(A \mid E_1) + P(E_2) \cdot P(A \mid E_2) + P(E_3) \cdot P(A \mid E_3) + P(E_4) \cdot P(A \mid E_4)$
$⇒\frac{1}{2} = \frac{1}{4} \times \frac{12}{100} + \frac{1}{4} \times \frac{60}{100} + \frac{1}{4} \times \frac{x}{100} + \frac{1}{4} \times \frac{84}{100}$
$ \Rightarrow \frac{1}{2} = \frac{12 + 60 + x + 84}{400}$
$\Rightarrow \frac{400}{2} = 156 + x$
$\Rightarrow x = 200 – 156$
$⇒x = 44$
$ P\left(\frac{A}{E_3}\right) = 44\%$