Out of 120 customers in a shop, 45 bought both bags and shoes. If all the customers bought either bags or shoes and 11 more customers bought shoes than bags...

Answer Details

Let \(b\) be the number who bought bags and \(s\) the number who bought shoes. Every customer bought at least one item, so \(n(\text{bags} \cup \text{shoes}) = 120\), with \(n(\text{both}) = 45\).

Using \(n(B \cup S) = b + s - n(\text{both})\):

\[120 = b + s - 45 \Rightarrow b + s = 165.\]

Also 11 more bought shoes than bags: \(s = b + 11\).

(a) Diagram. A two-circle Venn diagram (Bags and Shoes) with the overlap \(= 45\), bags-only \(= b - 45\), shoes-only \(= s - 45\).

(b) Substitute \(s = b + 11\) into \(b + s = 165\):

\[b + (b + 11) = 165 \Rightarrow 2b = 154 \Rightarrow b = 77, \quad s = 88.\]

The number who bought shoes is \(\mathbf{88}\).

(c) The number who bought bags is 77, so

\[P(\text{bought bags}) = \frac{77}{120}.\]

Checking the regions: bags-only \(= 32\), both \(= 45\), shoes-only \(= 43\); total \(= 120\).