c) Find the probability that the shopper buys milk and bread but not apples.
My attempt: $P(A' \cap B \cap C)=(1-.3) \times .45 \times .4=0.126$
$$P(A' \cap B \cap C)=\mathbf{0.18}\ne0.126= P(A')\times P(B)\times P(C).$$ (The correct answer $\mathbf{0.18}$ is explained below.) Thus, events $A',B,C$ are not in fact mutually independent. So, events $A,B,C$ are also not independent.
Incidentally, note that
- $P(X \cap Y \cap Z)=P(X)\times P(Y)\times P(Z){\kern.6em\not\kern-.6em\implies}$ $X,Y,Z$ are independent
- $X,Y,Z$ are pairwise independent ${\kern.6em\not\kern-.6em\implies} P(X \cap Y \cap Z)=P(X)\times P(Y)\times P(Z).$
b) Find the probability that the shopper purchases none of the three items.
My attempt: $P(A' \cup B' \cup C')=(1-.3) \times (1-.45) \times (1-.4)=.231$
You meant to write $P(A' \cap B' \cap C');$ unfortunately, as above, $$P(A' \cap B' \cap C')\ne P(A')\times P(B')\times P(C').$$
a) Find the probability that the shopper purchases at least one of the three items.
My attempt: $P(A \cup B \cup C) = P(A)+P(B)+P(C)-P(A \cap B)-P(B
\cap C)-P(A \cap C)+P(A \cap B \cap C)=.30+.45+.40-.2-.25-.12 + .17 =0.75$
You misread $0.07$ as $0.17,$ so your answer is too big by $0.10.$
A CORRECT SOLUTION
Let $A$ be the event that a randomly selected shopper buys apples, $B$ be the event that the same randomly selected shopper buys milk, and $C$ the event that the shopper buys bread.
Why does the problem statement make $B$ stand for Milk instead of Bread? To lower the cognitive load, thereby working faster and less carelessly, it would be better to begin by redefining the sets as $A,M,B.$ However, below, I'm just sticking to the assigned set names, $A,B,C.$
- $30\%$ of all shoppers buy apples
- $45\%$ buy milk
- $40\%$ buy a loaf of bread
- probability that buys apples and milk is $0.20$
- probability that buys milk and bread is $0.25$
- probability that buys apples and bread is $0.12$
- probability that buys all three items is $0.07$
A Venn diagram (with the probabilities scaled by $100)$ can capture the above information. Start from the last bullet point and work upwards.
c) Find the probability that the shopper buys milk and bread but not apples.
c) $\mathbf{0.18}$
b) Find the probability that the shopper purchases none of the three items.
b) $\mathbf{0.35}$
a) Find the probability that the shopper purchases at least one of the three items.
a) $1-0.35=\mathbf{0.65}$ (alternatively, add up the numbers in the circles then divide by $100).$
Best Answer
If she only buys one item, there are two possibilities: $A$ or $B$.
$$P(B|A \cup B) = \frac{P(B)}{P(A) + P(B)}$$
The case of purchasing two objects is irrelevant.
And you should have seen that your reasoning was wrong once you computed a probability of $1.8$.