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There are two balls in an urn. Each ball can be either white or black. If a white ball is put into the urn and there after a ball is drawn at random from the urn, then the probability that it is white is
$\frac{1}{4}$
$\frac{2}{3}$
$\frac{1}{5}$
$\frac{1}{3}$
Solution
We know that there are two ball in the urn and each can be either white or black ball so there will be three case,
$1) BB$
$2)WW$
$3)BW$
Now when we will put a white ball in the urn we will get our cases as ,
$1) BBW$
$2)WWW$
$3)BWW$
Now probability that the drawn ball is white from first case $=\frac{1}{3}$
Probability that the drawn ball is white from second case $=\frac{3}{3}=1$ [because all three balls are white]
Probability that the drawn ball is white from third case $=\frac{2}{3}$
Now all these three cases are independent of each other so
Required probability $\frac{1}{3} \times 1 \times \frac{2}{3} = \frac{2}{9}$
