Question

An urn contains 4 yellow balls and 6white balls. If Sam chooses 5 balls atrandom from the urn, what is theprobability that he will select 3 yellowballs and 2 white balls? Round to 3decimal places

Answer 1

Our approach is the binomial probability model.

Let p represent the probability of obtaining a yellow ball.

`p=\frac{\#\text{ of yellow balls}}{\#\text{ of total balls}}=(4)/(10)=(2)/(5)`

Let q represent the probability of obtaining a non-yellow ball, i.e, a white ball.

`p=\frac{\#\text{ of white balls}}{\#\text{ of total balls}}=(6)/(10)=(3)/(5)`

Binomial probability

To get r successes in n trial is:

`^nC_rp^rq^(n-r)`

where:

n = number of balls chosen at random

r = number of yellow balls to be selected

n- r = number of white balls to be selected

Applying, we get:

`\begin{gathered} ^5C_3\cdot p^3q^2 \\ =(5*4*3!)/(3!*2!)*((2)/(5))^3*((3)/(5))^2=0.2304 \end{gathered}`

The probability that he will select 3 yellow balls and 2 white balls is 0.230