Our approach is the binomial probability model.
Let p represent the probability of obtaining a yellow ball.
[tex]p=\frac{\#\text{ of yellow balls}}{\#\text{ of total balls}}=\frac{4}{10}=\frac{2}{5}[/tex]Let q represent the probability of obtaining a non-yellow ball, i.e, a white ball.
[tex]p=\frac{\#\text{ of white balls}}{\#\text{ of total balls}}=\frac{6}{10}=\frac{3}{5}[/tex]Binomial probability
To get r successes in n trial is:
[tex]^nC_rp^rq^{n-r}[/tex]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:
[tex]\begin{gathered} ^5C_3\cdot p^3q^2 \\ =\frac{5\times4\times3!}{3!\times2!}\times(\frac{2}{5})^3\times(\frac{3}{5})^2=0.2304 \end{gathered}[/tex]The probability that he will select 3 yellow balls and 2 white balls is 0.230
