Box 1 contains 30 cards numbered 1 to 30 and Box 2 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

Correct answer: B. (8)/(17) P( B _ 1 ) = (1)/(2) = P ( B _ 2 ) Box B _ 1 have 20 non-prime numbers \ 1, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30\ ⇒ P (NP / B _ 1 ) = (20)/(30) Box B _ 1 have 15 non-prime numbers \ 32, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48, 49, 50\ ⇒ P (NP / B _ 2 ) = (15)/(20) Probability of non-prime= P (NP). P(NP) = P ( B _ 1 ) . P (NP / B _ 1 ) + P ( B _ 2 ) . P (NP / B _ 2 ) =(1)/(2) · (20)/(30) + (1)/(2) · (15)/(20) P( B _ 1 / NP) = ((1)/(2) · (20)/(30))/((1)/(2) · (20)/(30) + (1)/(2) · (15)/(20)) = (8)/(17)

JEE Main 20202 SeptemberMorning Shift

Exam: JEE Main
Year: 2020
Date: 2 September
Shift: Morning Shift
Subject: Mathematics
Chapter: Probability

JEE Main 2020 · 2 September · Morning Shift · Mathematics · Probability

Box 1 contains 30 cards numbered 1 to 30 and Box 2 contains 20...

Box

1

contains

30

cards numbered

1

30

and Box

2

contains

20

cards numbered

31

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

1

Options

A $\dfrac{2}{3}$
B $\dfrac{8}{17}$
C $\dfrac{4}{17}$
D $\dfrac{2}{5}$

Correct answer

B. $\dfrac{8}{17}$

Step-by-step solution

$P\left({B}_{1}\right) = \dfrac{1}{2} =$ P \left({B}_{2}\right) $Box$ {B}_{1} $have$ 20 $non-prime numbers \left\{1, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30\right\}$

$⇒ P \left(NP / {B}_{1}\right) = \dfrac{20}{30}$

Box ${B}_{1}$ have $15$ non-prime numbers \left\{32, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48, 49, 50\right\} $$ ⇒ P \left(NP / {B}_{2}\right) = \dfrac{15}{20} $Probability of$ non-prime= P \left(NP\right).

$ $P\left(NP\right)$ = P \left({B}_{1}\right) . P \left(NP / {B}_{1}\right) + P \left({B}_{2}\right) . P \left(NP / {B}_{2}\right)

=\dfrac{1}{2} \cdot \dfrac{20}{30} + \dfrac{1}{2} \cdot \dfrac{15}{20} $$ P\left({B}_{1} / NP\right) = \dfrac{\dfrac{1}{2} \cdot \dfrac{20}{30}}{\dfrac{1}{2} \cdot \dfrac{20}{30} + \dfrac{1}{2} \cdot \dfrac{15}{20}} = \dfrac{8}{17}$

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