A nationwide survey showed that 65% of all children in Ghana dislike eating vegetables. If 4 children are chosen at random, what is the probability that all 4 dislike eating vegetables?
Consider these events independent because there are a lot more than 4 children in Ghana.
Use binomial distribution.
C(4,4)(.65)^4 (.35)^0
1 * .65^4 * 1
.65^4
note C(n,k) = n!/[k!(n-k)!]
defined as 1 for k = 0
see
http://www.mathsisfun.com/data/binomial-distribution.html
To find the probability that all four children dislike eating vegetables, we need to use the concept of probability and the given information.
First, let's define some terms:
- P(A): Probability of children disliking eating vegetables
- P(A'): Probability of children liking eating vegetables
Given information:
The nationwide survey showed that 65% of all children in Ghana dislike eating vegetables. This means that the probability of a child disliking eating vegetables is P(A) = 0.65 (or 65%).
Since each child's preference is independent of one another, we can use the multiplication rule for independent events. According to this rule, the probability of two independent events occurring together is the product of their individual probabilities.
For the first child, the probability of them disliking eating vegetables is P(A) = 0.65.
For the second child, the probability is also P(A) = 0.65.
For the third child, the probability is once again P(A) = 0.65.
For the fourth child, the probability is P(A) = 0.65.
Using the multiplication rule, the probability that all four children dislike eating vegetables is:
P(all four dislike vegetables) = P(A) * P(A) * P(A) * P(A)
= 0.65 * 0.65 * 0.65 * 0.65
Simplifying this expression, we get:
P(all four dislike vegetables) = 0.65^4
Therefore, the probability that all four children chosen at random dislike eating vegetables is 0.65^4, which is approximately 0.179 or 17.9%.