Answer:0.375
Explanation:
Given
An Urn contains 3 White and 3 black balls
Ball is replaced after it is drawn
Using Binomial Distribution as trials are finite
n=4 i.e. 4 balls are drawn
Probability of getting white ball [tex](p)=\frac{1}{2}[/tex]
Probability of getting a Non-white ball[tex](q)=\frac{1}{2}[/tex]
[tex]P(X=r)= ^nC_r(p)^r(q)^{n-r}[/tex]
For Exactly 2 white balls
[tex]P(X=2)=^4C_2(\frac{1}{2})^{2}(\frac{1}{2})^{2}[/tex]
[tex]P(X=2)=\frac{4!}{2!\cdot 2!}\times \frac{1}{2^4}[/tex]
[tex]P(X=2)=\frac{3}{8}[/tex]
