There are two balls in an urn. Each ball can | vedclass

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Question

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,

$1) BB$

$2)WW$

$3)BW$

N

Vedclass · Accepted Answer

$\frac{2}{3}$

Vedclass · Answer

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 $\frac{1}{3} 	imes 1 	imes \frac{2}{3} = \frac{2}{9}$

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

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