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JEE Main 20266 AprilMorning Shift
Exam
JEE Main
Year
2026
Date
6 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Probability

JEE Main 2026 · 6 April · Morning Shift · Mathematics · Probability

A bag contains (N+1) coins - N fair coins, and one coin with 'Head'...

A bag contains (N+1)(N+1) coins −N- N fair coins, and one coin with 'Head' on both sides. A coin is selected at random and tossed. If the probability of getting 'Head' is 916\dfrac{9}{16}, then NN is equal to:

Options

  1. A55
  2. B77
  3. C88
  4. D99

Correct answer

B. 77

Step-by-step solution

Let E1E_1 be the event of selecting a fair coin and E2E_2 be the event of selecting the two-headed coin.

P(E1)=NN+1P(E_1) = \dfrac{N}{N+1}

P(E2)=1N+1P(E_2) = \dfrac{1}{N+1}

Let HH be the event of getting a Head.

P(H∣E1)=12P(H|E_1) = \dfrac{1}{2}

P(H∣E2)=1P(H|E_2) = 1

Using the theorem of total probability:

P(H)=P(E1)P(H∣E1)+P(E2)P(H∣E2)P(H) = P(E_1)P(H|E_1) + P(E_2)P(H|E_2)

P(H)=(NN+1)(12)+(1N+1)(1)P(H) = \left(\dfrac{N}{N+1}\right)\left(\dfrac{1}{2}\right) + \left(\dfrac{1}{N+1}\right)(1)

P(H)=N+22(N+1)P(H) = \dfrac{N+2}{2(N+1)}

Given P(H)=916P(H) = \dfrac{9}{16}:

N+22(N+1)=916\dfrac{N+2}{2(N+1)} = \dfrac{9}{16}

N+2N+1=98\dfrac{N+2}{N+1} = \dfrac{9}{8}

8N+16=9N+98N + 16 = 9N + 9

N=7N = 7

Answer: 77
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