Question

A bag contains $5$ red and $3$ blue balls. If $3$ balls are drawn at random without replacement then the probability that exactly two of three balls were red, the first ball being red, is

Answer 1

Given number of red balls = $5$
Number of blue balls = $3$
Total number of balls = $8$
$\therefore$ Probabilty of any event is :
$p\left(A\right)=\frac{\text{No.of favourable outcomes w.r.t.A}}{\text{Total outcomes of the experiment}}$
$\therefore$ The probability that exactly two of three balls were red, the first ball being red, is Required probability $=P\left(R\right)\cdot P\left(B\right)+P\left(B\right)\cdot P\left(R\right)=\frac{4}{7}\cdot \frac{3}{6}+\frac{3}{7}\cdot \frac{4}{6}$
[$\therefore$ As First ball is red]
$=\frac{4}{7}$
$\therefore$ The probability that exactly two of the three balls were red, the first ball being red is
$\frac{4}{7}$
Hence option (B) is correct.