JEE Advance - Mathematics (2019 - Paper 1 Offline - No. 10)
There are three bags B1, B2 and B3. The bag B1 contains 5 red and 5 green balls, B2 contains 3 red and 5 green balls, and B3 contains 5 red and 3 green balls. Bags B1, B2 and B3 have probabilities $${3 \over {10}}$$, $${3 \over {10}}$$ and $${4 \over {10}}$$ respectively of being chosen. A bag is selected at random and a ball is chosen at random from the bag. Then which of the following options is/are correct?
Probability that the chosen ball is green, given that the selected bag is B3, equals $${3 \over 8}$$.
Probability that the selected bag is B3, given that the chosen ball is green, equals $${5 \over 13}$$.
Probability that the chosen ball is green equals $${39 \over 80}$$.
Probability that the selected bag is B3 and the chosen ball is green equals $${3 \over 10}$$.
Explanation
It is given that there are three bags B1, B2 and B3 and probabilities of being chosen B1, B2 and B3 are respectively.
$$ \therefore $$ $$P({B_1}) = {3 \over {10}},\,P({B_2}) = {3 \over {10}}$$ and $$P({B_3}) = {4 \over {10}}$$.
Now, probability that the chosen ball is green, given that selected bag is
$${B_3} = P\left( {{G \over {{B_3}}}} \right) = {3 \over 8}$$
Now, probability that the selected bag is B3, given that the chosen ball is green = $$P\left( {{{{B_3}} \over G}} \right)$$
$$ = {{P\left( {{G \over {{B_3}}}} \right)P({B_3})} \over {P\left( {{G \over {{B_1}}}} \right)P({B_1}) + P\left( {{G \over {{B_2}}}} \right)P({B_2}) + P\left( {{G \over {{B_3}}}} \right)P({B_3})}}$$
[by Baye's theorem]
= $${{\left( {{3 \over 8} \times {4 \over {10}}} \right)} \over {\left( {{5 \over {10}} \times {3 \over {10}}} \right) + \left( {{5 \over 8} \times {3 \over {10}}} \right) + \left( {{3 \over 8} \times {4 \over {10}}} \right)}}$$
$$ = {{{1 \over 2}} \over {{1 \over 2} + {5 \over 8} + {1 \over 2}}} = {4 \over {13}}$$
Now, probability that the chosen ball is green = P(G)
= $$P({B_1})P\left( {{G \over {{B_1}}}} \right) + P({B_2})P\left( {{G \over {{B_2}}}} \right) + P({B_3})P\left( {{G \over {{B_3}}}} \right)$$
[By using theorem of total probability]
= $$\left( {{3 \over {10}} \times {5 \over {10}}} \right) + \left( {{3 \over {10}} \times {5 \over 8}} \right) + \left( {{4 \over {10}} \times {3 \over 8}} \right)$$
= $${3 \over {20}} + {3 \over {16}} + {3 \over {20}} = {{12 + 15 + 12} \over {80}} = {{39} \over {80}}$$
Now, probability that the selected bag is B3 and the chosen ball is green
= $$P({B_3}) \times P\left( {{G \over {{B_3}}}} \right) = {4 \over {10}} \times {3 \over 8} = {3 \over {20}}$$
Hence, options (a) and (c) are correct.
$$ \therefore $$ $$P({B_1}) = {3 \over {10}},\,P({B_2}) = {3 \over {10}}$$ and $$P({B_3}) = {4 \over {10}}$$.
Now, probability that the chosen ball is green, given that selected bag is
$${B_3} = P\left( {{G \over {{B_3}}}} \right) = {3 \over 8}$$
Now, probability that the selected bag is B3, given that the chosen ball is green = $$P\left( {{{{B_3}} \over G}} \right)$$
$$ = {{P\left( {{G \over {{B_3}}}} \right)P({B_3})} \over {P\left( {{G \over {{B_1}}}} \right)P({B_1}) + P\left( {{G \over {{B_2}}}} \right)P({B_2}) + P\left( {{G \over {{B_3}}}} \right)P({B_3})}}$$
[by Baye's theorem]
= $${{\left( {{3 \over 8} \times {4 \over {10}}} \right)} \over {\left( {{5 \over {10}} \times {3 \over {10}}} \right) + \left( {{5 \over 8} \times {3 \over {10}}} \right) + \left( {{3 \over 8} \times {4 \over {10}}} \right)}}$$
$$ = {{{1 \over 2}} \over {{1 \over 2} + {5 \over 8} + {1 \over 2}}} = {4 \over {13}}$$
Now, probability that the chosen ball is green = P(G)
= $$P({B_1})P\left( {{G \over {{B_1}}}} \right) + P({B_2})P\left( {{G \over {{B_2}}}} \right) + P({B_3})P\left( {{G \over {{B_3}}}} \right)$$
[By using theorem of total probability]
= $$\left( {{3 \over {10}} \times {5 \over {10}}} \right) + \left( {{3 \over {10}} \times {5 \over 8}} \right) + \left( {{4 \over {10}} \times {3 \over 8}} \right)$$
= $${3 \over {20}} + {3 \over {16}} + {3 \over {20}} = {{12 + 15 + 12} \over {80}} = {{39} \over {80}}$$
Now, probability that the selected bag is B3 and the chosen ball is green
= $$P({B_3}) \times P\left( {{G \over {{B_3}}}} \right) = {4 \over {10}} \times {3 \over 8} = {3 \over {20}}$$
Hence, options (a) and (c) are correct.
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