Verified Solution • 2025–26 Session moderate

QJEEAdv19Ma0eca: JEE Advanced 2019 Mathematics - Class 12 JEEAdv19Ma0eca Solution: There are three bags B 1 , B 2 and B 3 . The bag B 1 contains 5 …

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 310{3 \over {10}} , 310{3 \over {10}} and 410{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?

Multiple Choice Options

  • A Probability that the chosen ball is green, given that the selected bag is B3, equals 38{3 \over 8} .
  • B Probability that the selected bag is B3, given that the chosen ball is green, equals 513{5 \over 13} .
  • C Probability that the chosen ball is green equals 3980{39 \over 80} .
  • D Probability that the selected bag is B3 and the chosen ball is green equals 310{3 \over 10} .

Step-by-Step Verified Solution

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(B1)=310, P(B2)=310P({B_1}) = {3 \over {10}},\,P({B_2}) = {3 \over {10}} and P(B3)=410P({B_3}) = {4 \over {10}} .

JEE Advanced 2019 Paper 1 Offline Mathematics - Probability Question 26 English Explanation

Now, probability that the chosen ball is green, given that selected bag is

B3=P(GB3)=38{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(B3G)P\left( {{{{B_3}} \over G}} \right)

=P(GB3)P(B3)P(GB1)P(B1)+P(GB2)P(B2)+P(GB3)P(B3)= {{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]

= (38×410)(510×310)+(58×310)+(38×410){{\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)}}

=1212+58+12=413= {{{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(B1)P(GB1)+P(B2)P(GB2)+P(B3)P(GB3)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]

= (310×510)+(310×58)+(410×38)\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)

= 320+316+320=12+15+1280=3980{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(B3)×P(GB3)=410×38=320P({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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