Verified Solution • 2025–26 Session moderate

QJEEM20Ma0a78: JEE Main 2020 Mathematics - Class 12 JEEM20Ma0a78 Solution: Box I contains 30 cards numbered 1 to 30 and Box II contains 20 …

Box I contains 30 cards numbered 1 to 30 and Box II contains 20 cards numbered 31 to 50. A box is selected at random and a card is drawn from it. The number on the card is found to be a non-prime number. The probability that the card was drawn from Box I is :

Multiple Choice Options

  • A 817{8 \over {17}}
  • B 23{2 \over 3}
  • C 25{2 \over 5}
  • D 417{4 \over {17}}

Step-by-Step Verified Solution

Let B1 be the event where Box-I is selected.

And B2 be the event where Box-II is selected.

P(B1) = P(B2) = 12{1 \over 2}

Let E be the event where selected card is non prime.

For B1 : Prime numbers: {2, 3, 5, 7, 11, 13, 17, 19, 23, 29}

For B2 : Prime numbers: {31, 37, 41, 43, 47}

P(E) = P(B1) ×\times P(EB1)P\left( {{E \over {{B_1}}}} \right) + P(B2) ×\times P(EB2)P\left( {{E \over {{B_2}}}} \right)

= 12×2030{1 \over 2} \times {{20} \over {30}} + 12×1520{1 \over 2} \times {{15} \over {20}}

Required probability :

P(B1E)P\left( {{{{B_1}} \over E}} \right) = P(B2).P(EB1)P(E){{P\left( {{B_2}} \right).P\left( {{E \over {{B_1}}}} \right)} \over {P\left( E \right)}}

= 12×203012×2030+121520{{{1 \over 2} \times {{20} \over {30}}} \over {{1 \over 2} \times {{20} \over {30}} + {1 \over 2}{{15} \over {20}}}}

= 2323+34{{{2 \over 3}} \over {{2 \over 3} + {3 \over 4}}}

= 817{8 \over {17}}

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