Question

A bag contains 3 red and 7 black balls. Two balls are selected at random one-by-one without replacement. The probability that the second selected ball happens to be red if the first selected ball is red, is $\frac{2}{k}.$ Find k.

Answer 1

A bag contains $3$ red and $7$ black balls.

Therefore, total number of balls in the bag is $10$. That is $n\left(S\right)=10$.

Two balls are selected at random one-by-one without replacement.

The probability that the second selected ball happens to be red if the first selected ball is red, is $\frac{2}{k}$. We have to find the value of $k$.

Let $A$ be the event of selecting first ball as red.

Therefore, $P\left(A\right)=\frac{3}{10}$.

Without replacement of the ball, let $B$ be the event of selecting second ball to be red.

Therefore, $P\left(B\right)=\frac{2}{9}$

The probability that the second selected ball happens to be red if the first selected ball is red $=\frac{2}{k}$

$\Rightarrow P\left(B/A\right)=\frac{2}{k}$

$\Rightarrow \frac{P(B\cap A)}{P\left(A\right)}=\frac{2}{k}$

$\Rightarrow \frac{P\left(B\right)\cdot P\left(A\right)}{P\left(A\right)}=\frac{2}{k}$

$\Rightarrow \frac{\frac{2}{9}\cdot \frac{3}{10}}{\frac{3}{10}}=\frac{2}{k}$

$\Rightarrow \frac{2}{9}=\frac{2}{k}$

$\Rightarrow k=9$

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