Unions & Intersection of Probability

Question

I've had all three answers below marked wrong, and I am not sure how to proceed. I have included my thinking.

Suppose we are interested in the buying habits of shoppers at a particular grocery store with regards to whether they purchase apples, milk, and/or bread. Now suppose $30\%$ of all shoppers at this particular grocery store buy apples, $45\%$ buy milk, and $40\%$ buy a loaf of bread.
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. Suppose we also know (from data collected) the following information:

• The probability that the shopper buys apples and milk is $0.20$.
• The probability that the shopper buys milk and bread is $0.25$.
• The probability that the shopper buys apples and bread is $0.12$.
• The probability that the shopper buys all three items is $0.07$.

a) Find the probability that the shopper purchases at least one of the three items.

My attempt :
$P(A \cup B \cup C) =
P(A)+P(B)+P(C)-P(A \cap B)-P(B \cap C)-P(A \cap C)+P(A \cap B \cap C)=.30+.45+.40-.2-.25-.12 + .17 =0.75$

b) Find the probability that the shopper purchases none of the three items.

My attempt :
$P(A' \cup B' \cup C')=(1-.3) \times (1-.45) \times (1-.4)=.7 \times .45 \times .4 = .231$

c) Find the probability that the shopper buys milk and bread but not apples.

My attempt :
$P(A' \cap B \cap C)=(1-.3) \times .45 \times .4 =0.126$

EDIT

I have also tried doing this with naive probability and the answers are still not accepted.

a)
s=c(111, 110, 101, 011, 100, 001, 010, 000)
s_count=8
at_least_1= c(111, 110, 101, 011, 100, 001, 010)
at_least_1_ct=7
7/8=0.875

b)
none=c(111, 110, 101, 011, 100, 001, 010, 000)
1/8=0.125

c)
milk_and_bread=c(111, 110, 101, 011, 100, 001, 010, 000)
2/8=0.25

Answer 1

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 $0.10.$

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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 $100).$