We are given that out of 120 customers, 45 bought both bags and shoes. All customers bought either bags or shoes. Also, 11 more customers bought shoes than bags. We need to (a) illustrate this information in a Venn diagram, (b) find the number of customers who bought shoes, and (c) calculate the probability that a randomly selected customer bought bags.

1. Problem Description

We are given that out of 120 customers, 45 bought both bags and shoes. All customers bought either bags or shoes. Also, 11 more customers bought shoes than bags. We need to (a) illustrate this information in a Venn diagram, (b) find the number of customers who bought shoes, and (c) calculate the probability that a randomly selected customer bought bags.

2. Solution Steps

(a) Venn Diagram Illustration:

Let $B$ be the number of customers who bought bags, and $S$ be the number of customers who bought shoes.

We are given that the total number of customers is

1

2

0. Thus, $n(B \cup S) = 120$.

We are also given that the number of customers who bought both bags and shoes is

4

5. Thus, $n(B \cap S) = 45$.

Also, it is given that the number of customers who bought shoes is 11 more than the number of customers who bought bags. Thus, $n(S) = n(B) + 11$.

The Venn diagram has two circles representing bags and shoes. The intersection of the circles represents customers who bought both, which is

4

5. Let the number of customers who bought only bags be $x$, and the number of customers who bought only shoes be $y$. Then, $n(B) = x + 45$ and $n(S) = y + 45$.

We have $x + y + 45 = 120$.

From $n(S) = n(B) + 11$, we get $y + 45 = x + 45 + 11$, which implies $y = x + 11$.

(b) Number of customers who bought shoes:

Substitute $y = x + 11$ into $x + y + 45 = 120$.

$x + (x + 11) + 45 = 120$

$2x + 56 = 120$

$2x = 120 - 56 = 64$

$x = 32$

So, $n(\text{only bags}) = 32$.

Then $y = x + 11 = 32 + 11 = 43$.

So, $n(\text{only shoes}) = 43$.

The number of customers who bought shoes is $n(S) = y + 45 = 43 + 45 = 88$.

(c) Probability that a customer selected at random bought bags:

$n(B) = x + 45 = 32 + 45 = 77$.

The probability that a randomly selected customer bought bags is $P(B) = \frac{n(B)}{\text{Total customers}} = \frac{77}{120}$.

3. Final Answer

(a) The Venn diagram consists of two circles representing bags and shoes. The intersection of the circles has

4

5. The number of customers who bought only bags is 32, and the number of customers who bought only shoes is

4

3. (b) The number of customers who bought shoes is

8

8. (c) The probability that a customer selected at random bought bags is $\frac{77}{120}$.