Probability: Question 6

Worked solution

Step 1: Find the total number of badminton players

From the table, the number of students who play badminton is those who play “badminton only” plus those who play “both”: $n(\text{badminton}) = 14 + 8 = 22$

Step 2: Identify the number who play both badminton and chess

From the table, the number who play both is given directly: $n(\text{badminton and chess}) = 8$

Step 3: Apply the conditional probability formula

$P(\text{chess} \mid \text{badminton}) = \frac{P(\text{chess and badminton})}{P(\text{badminton})} = \frac{n(\text{badminton and chess})}{n(\text{badminton})} = \frac{8}{22} = \frac{4}{11}$

Check by working with probabilities directly

Using probabilities out of the full class of $40$: $P(\text{badminton}) = \frac{22}{40} = 0.55$ and $P(\text{chess and badminton}) = \frac{8}{40} = 0.2$. So $P(\text{chess} \mid \text{badminton}) = \frac{0.2}{0.55} = \frac{4}{11}$ which agrees exactly with Step 3.

Why the other options are wrong

• B ($\frac{1}{5}$): this is $\frac{8}{40}$, the probability of playing both out of the whole class, not restricted to badminton players.
• C ($\frac{4}{7}$): this is $\frac{8}{14}$, the reversed conditional probability $P(\text{badminton} \mid \text{chess})$.
• D ($\frac{7}{11}$): this is $\frac{14}{22}$, the probability of playing badminton only (not chess) given badminton, i.e. $1 - \frac{4}{11}$.

Final answer

• $P(\text{chess} \mid \text{badminton}) = \boxed{\dfrac{4}{11}}$, option A.