Probability: Question 10

Worked solution

Part (a): Reading probabilities from the table

From the table, $n(X) = 54+30=84$ (the ”$X$ only” and “both” regions), $n(Y)=66+30=96$, and $n(X \cap Y) = 30$ directly. Out of the $200$ subscribers: $P(X) = \frac{84}{200} = 0.42 \qquad P(Y) = \frac{96}{200} = 0.48 \qquad P(X \cap Y) = \frac{30}{200} = 0.15$

Part (b): Addition rule and the complement

$P(X \cup Y) = P(X) + P(Y) - P(X \cap Y) = 0.42 + 0.48 - 0.15 = 0.75$

“Reads neither” is the complement of “reads $X$ or $Y$”: $P(\text{neither}) = 1 - P(X \cup Y) = 1 - 0.75 = 0.25$

Check directly from the table: the “neither” region has $50$ subscribers, so $P(\text{neither}) = \frac{50}{200} = 0.25$, which agrees exactly.

Part (c): Testing independence

Two events are independent exactly when $P(X \cap Y) = P(X) \times P(Y)$. $P(X) \times P(Y) = 0.42 \times 0.48 = 0.2016$

Since $P(X \cap Y) = 0.15 \neq 0.2016$, the events “reads $X$” and “reads $Y$” are not independent. A subscriber who reads $X$ is in fact less likely to also read $Y$ than under independence.

Part (d): Conditional probability given the complement

First find $P(Y')$, the probability of not reading $Y$: $P(Y') = 1 - P(Y) = 1 - 0.48 = 0.52$

The event ”$X$ and not $Y$” corresponds exactly to the ”$X$ only” region: $P(X \cap Y') = \frac{54}{200} = 0.27$

So $P(X \mid Y') = \frac{P(X \cap Y')}{P(Y')} = \frac{0.27}{0.52} = \frac{27}{52} \approx 0.519$

Check directly from the table: subscribers who do not read $Y$ number $54 + 50 = 104$ (the ”$X$ only” plus “neither” regions), of whom $54$ read $X$: $P(X \mid Y') = \frac{54}{104} = \frac{27}{52} \approx 0.519$ which agrees exactly with the calculation above.

Final answers

• (a) $P(X) = \boxed{0.42}$, $P(Y) = \boxed{0.48}$, $P(X \cap Y) = \boxed{0.15}$
• (b) $P(X \cup Y) = \boxed{0.75}$; $P(\text{neither}) = \boxed{0.25}$
• (c) Not independent, since $0.2016 \neq 0.15$.
• (d) $P(X \mid Y') = \boxed{\dfrac{27}{52}} \approx 0.519$