JEE Advanced 2023 Paper 2 · Mathematics · Probability · medium

JEE Advanced 2023 Paper 2 Mathematics: Paragraph 2

A fair six-faced die is rolled once. If the number obtained is

k

, then

k

fair coins are tossed. Let

E

be the event that exactly two heads appear in these tosses.

Question 1

Numerical answer

The probability

P(E)

is (give the answer correct to two decimal places)

Show answer and solution

Answer: 0.26

P(E)=\sum_{k=1}^{6}\frac16\binom k2\frac1{2^k}

. For

k=1

the term is

0

; for

k=2,\dots,6

the values of

\binom k2/2^k

are

\frac14,\frac38,\frac38,\frac5{16},\frac{15}{64}

, whose sum is

\frac{16+24+24+20+15}{64}=\frac{99}{64}

. Hence

P(E)=\frac16\cdot\frac{99}{64}=\frac{33}{128}\approx0.2578

, which is

0.26

to two decimal places.

Numerical answer

Given that

E

has occurred, the probability that the die showed

4

is (give the answer correct to two decimal places)

Show answer and solution

Answer: 0.24

P(4\mid E)=\frac{P(4)P(E\mid4)}{P(E)}=\frac{\frac16\cdot\frac{6}{16}}{33/128}=\frac{1/16}{33/128}=\frac{8}{33}\approx0.24

Original question written in the JEE Advanced 2023 Paper 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake