JEE Advanced 2024 Paper 2 · Mathematics · Probability · medium

JEE Advanced 2024 Paper 2 Mathematics: Paragraph 2

A fair coin is tossed repeatedly until two consecutive heads appear for the first time. Let

X

denote the number of tosses required. It is known that the number of head-tail sequences of length

n\ge2

that end in

\mathrm{HH}

(for

n\ge3

, in

\mathrm{THH}

) and contain no earlier occurrence of two consecutive heads is the Fibonacci number

F_{n-1}

, where

F_1=F_2=1

F_3=2

, and so on. Give each answer correct to two decimal places.

Question 1

Numerical answer

The probability

P(X=6)

Show answer and solution

Answer: 0.08

With

P(X=n)=F_{n-1}/2^n

we have

P(X=6)=F_5/2^6=\dfrac{5}{64}=0.078125\approx0.08

. Equivalently, a valid length-6 sequence must end in THH, and its first three tosses must contain no HH; there are exactly

F_5=5

such three-toss prefixes (TTT, TTH, THT, HTT, HTH), each followed by THH, giving 5 sequences out of

2^6=64

Numerical answer

The probability

P(X\le6)

Show answer and solution

Answer: 0.67

Using

P(X=n)=F_{n-1}/2^n

P(2)=\tfrac14

P(3)=\tfrac18

P(4)=\tfrac{2}{16}

P(5)=\tfrac{3}{32}

P(6)=\tfrac{5}{64}

. Over the common denominator

64

these are

16,8,8,6,5

, summing to

\dfrac{43}{64}=0.671875\approx0.67

Original question written in the JEE Advanced 2024 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