Question

A bag contains $4$ green and $6$ white balls. Two balls are drawn one by one without replacement. If the second ball drawn is white, what is the probability that the first ball drawn is also white?

Answer 1

The first ball drawn is white: A

Second ball is white: B

conditionally probability :$P\left(A|B\right)$

$P\left(A|B\right)=P\left(AnB\right)/P\left(B\right)$ (I)

AnB= first ball was white and second ball is also white.

$P\left(AnB\right)=6/10\ast 5/9$ (II)

second ball is white: B

it can happen in two mutually exclusive ways :

a) first ball is white and second is white

b) first ball is green while the second is white.

$P\left(B\right)=P\left(1\right)+P\left(2\right)=6/10\ast 5/9+4/10\ast 6/9$ (iii)

Substituting equation (II) and (iii) in equation (I)

$P\left(A|B\right)$

$=(6/10\ast 5/9)/(6/10\ast 5/9+4/10\ast 6/9)$

$=30/(30+24)$

$=30/54$

=$\frac{5}{9}$