Question

A bag contains $3$ red and $7$ black balls. Two balls are drawn one by one at a time at random without replacement. If second drawn ball is red then what is the probability the first drawn ball is also red?

Answer 1

Number of red balls $=3$

Number of black balls $=7$

Total number of balls $=10$

Let ,$P\left(A\right)=$ Probability that first is red

$P\left(B\right)=$ Probability that second is red

If second is red, there are 2 possible ways

Either first is red and second is red.

Or first is black and second is red.

So,

$P\left(B\right)=\frac{{}^3{C}_1}{10}\times \frac{{}^2{C}_1}{9}+\frac{{}^7{C}_1}{10}\times \frac{{}^3{C}_1}{9}$

$=\frac{6}{90}+\frac{21}{90}$

$=\frac{27}{90}$

$=\frac{3}{10}$

Probability that first is red, given second is red

$P\left(A/B\right)=\frac{P(A\cap B)}{P\left(B\right)}$

$=\frac{\frac{{}^3{C}_1}{10}\times \frac{{}^2{C}_1}{9}}{\frac{3}{10}}$

$=\frac{\frac{6}{90}}{\frac{3}{10}}$

$=\frac{6}{90}\times \frac{10}{3}$

$=\frac{6}{27}$

$=\frac{2}{9}$