# Solved (Free): A supermarket collected data to estimate the proportion of shoppers that buy certain types of products

#### ByDr. Raju Chaudhari

Mar 14, 2021

A supermarket collected data to estimate the proportion of shoppers that buy certain types of products. It was found that 50% of shoppers bought dairy products, 30% bought meat, and 40% bought neither.

A shopper is selected at random. Let $A$ be the event they "bought dairy products", and let $B$ be the event they "bought meat". Let $\overline{A}$ and $\overline{B}$ denote the complements of $A$ and $B$, respectively.

i) Fill in the probabilities in a copy of the following two-way table:

. $B$ $\overline{B}$ Total
$A$
$\overline{A}$
Total 1

ii) Find the probability of the event "bought dairy products or meat"
iii) Find the probability that a shopper buys dairy products given they do not buy meat.
iv) Are events A and B mutually exclusive? Explain clearly why or why not.
v) Are events A and B independent? Explain clearly why or why not.

#### Solution

i) Given that $P(A) = 0.50$, so $P(\overline{A}) = 1- P(A) = 0.50$.

$P(B) = 0.30$, so $P(\overline{B}) = 1- P(B) = 0.70$.

$P(\overline{A}\cap \overline{B}) = 0.40$.

Using all these values and calculating remaining probabilities we have

. $B$ $\overline{B}$ Total
$A$ 0.20 0.30 0.50
$\overline{A}$ 0.10 0.40 0.50
Total 0.30 0.70 1

ii) The probability of the event "bought dairy products or meat" is

 \begin{aligned} P(A\cup B) &= P(A)+P(B) -P(A\cap B)\\ &= 0.50 +0.30 - 0.20 \\ &= 0.6. \end{aligned}

iii) The probability that a shopper buys dairy products given they do not buy meat is

 \begin{aligned} P(A|\overline{B}) &= \frac{P(A\cap \overline{B})}{P(\overline{B})}\\ &= \frac{0.30}{0.70}\\ &= 0.4286. \end{aligned}

iv) Events $A$ and $B$ are not mutually exclusive, because $P(A\cap B) = 0.20 \neq 0$.

v) $P(A\cap B) = 0.20$, $P(A) =0.50$, $P(B) = 0.30$.
So $P(A)\times P(B) = 0.50*0.30 = 0.15$.

Events $A$ and $B$ are not independent because $P(A\cap B) \neq P(A)\times P(B)$.