There are $3$ bags, each containing $5$ white balls and $3$ black balls. Also there are $2$ bags, each containing $2$ white balls and $4$ black balls. A white ball is drawn at random. If probability that this white ball is from a bag of the first group is $k$ , then the value of $61 k - 40$ is

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Solution

$E_{1} \to$ Bag from the first group

$E_{2} \to$ bag from the second group

$W \to$ drawn ball is white

$P (E_{1}) = \frac{3}{5}$ , $P (E_{2}) = \frac{2}{5}$ , $P (\frac{W}{E_{1}}) = \frac{5}{8}$ , $P (\frac{W}{E_{2}}) = \frac{2}{6} = \frac{1}{3}$

$P (\frac{E_{1}}{W}) = \frac{P (E_{1}) . P (\frac{W}{E_{1}})}{P (E_{1}) P (\frac{W}{E_{1}}) + P (E_{2}) . P (\frac{W}{E_{2}})}$

$= \frac{\frac{3}{5} \times \frac{5}{8}}{\frac{3}{5} \times \frac{5}{8} + \frac{2}{5} \times \frac{1}{3}} = \frac{45}{61}$

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There are 3 bags, each containing 5 white balls and 3 black balls. Also there are 2 bags, each containing 2 white balls and 4 black balls. A white ball is drawn at random. If probability that this white ball is from a bag of the first group is k, then the value of 61k−40 is

E1→ Bag from the first group E2→ bag from the second group W→ drawn ball is white P(E1)=35, P(E2)=25, P(WE1)=58, P(WE2)=26=13 P(E1W)=P(E1).P(WE1)P(E1)P(WE1)+P(E2).P(WE2) =35×5835×58+25×13=4561

There are $3$ bags, each containing $5$ white balls and $3$ black balls. Also there are $2$ bags, each containing $2$ white balls and $4$ black balls. A white ball is drawn at random. If probability that this white ball is from a bag of the first group is $k$ , then the value of $61 k - 40$ is