Question

Suppose you know 65% of students of like vegetable, 71% students of like ham, and 48% of students of like both vegetable and ham. If you sample a student at random, what is probability they do not like both ham and vegetable? I want answer with step by step.

Answer 1

Step-by-step explanation

We can solve the question using the formula below

`\begin{gathered} P\left(H∪V\right)=P\left(H\right)+P\left(V\right)−P\left(H∩V\right) \\ Therefore; \\ P\left(H\cup V\right)^(\prime)=1-P\left(H∪V\right) \end{gathered}`

Therefore;

The probabilty of those that eat one or more of both foods

`\begin{gathered} P\left(H∪V\right)=P\left(H\right)+P\lparen V)−P\left(H∩V\right) \\ P\left(H∪V\right)=0.71+0.65-0.48 \\ P\left(H∪V\right)=0.88 \end{gathered}`

Hence, we can find the probability of those that do not like both ham and vegetable

`\begin{gathered} P(H\cup V)^(\prime)=1-P(H\cup V) \\ =1-0.88=0.12 \end{gathered}`

Answer: 0.12