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.

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.

$44\%$

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\%$