Given:
• Number of people who like only bananas = 12
,• Number of people who like only pies = 11
,• Number of people who like both = 7
,• Number of people who like neither = 3
Let's solve for the following:
• b(i). Given that a person is chosen at random, let's find the probability that the person likes banana.
To find the probability, apply the formula:
[tex]P(banana)=\frac{number\text{ of people who like banana}}{total\text{ number of people surveyed}}[/tex]Where:
Total number of people surveyed = 12 + 11 + 7 + 3 = 33
Number of people who like banana = 12 + 7= 19
Thus, we have:
[tex]P(bananas)=\frac{19}{33}[/tex]Therefore, the probability that a person chosen randomly likes bananas is 19/33.
• b(ii). ,Given that a person is chosen at random, let's find the probability that the person doesn't like pie.
To find the probability, apply the formula:
[tex]\begin{gathered} P(doesn^{\prime}t\text{ like pie\rparen = 1-}\frac{}{}P(people\text{ who like pie\rparen} \\ \\ P(doesn^{\prime}t\text{ like pie\rparen = 1-}\frac{number\text{ of people who like pie}}{total\text{ number of people}} \end{gathered}[/tex]Where:
Number of people who like pie = 11 + 7 = 18
Total number of people = 33
Thus, we have:
[tex]\begin{gathered} P(doesn^{\prime}t\text{ like pie\rparen = 1-}\frac{18}{33} \\ \\ =\frac{33-18}{33} \\ \\ =\frac{15}{33} \end{gathered}[/tex]The probability that a person chosen at random doesn't like pie is 15/33.
ANSWER:
• b(i). 19/33
• b(ii). 15/33