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Maths Question 10 – JEE-MAIN 2026

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A bag contains (N+1) coins – 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, then N is equal to:

There are two types of coins: fair coins and a two-headed coin. Determine the probability of selecting each type of coin.

Step 1: Define Events and Probabilities✦ Active

Let N be the number of fair coins. The total number of coins in the bag is N+1. There is 1 two-headed coin. The probability of selecting a fair coin is P(Fair)=NN+1. The probability of selecting the two-headed coin is P(Two-headed)=1N+1.

The probability of getting a head given a fair coin is P(Head|Fair)=12. The probability of getting a head given a two-headed coin is P(Head|Two-headed)=1.

Step 2: Apply the Law of Total Probability○ Expand

Using the Law of Total Probability, the overall probability of getting a head is:

P(Head)=P(Head|Fair)P(Fair)+P(Head|Two-headed)P(Two-headed)
P(Head)=(12)(NN+1)+(1)(1N+1)
P(Head)=N2(N+1)+22(N+1)=N+22(N+1)
💡 Teacher's Secret Hint

Ensure you correctly combine the fractions with a common denominator.

Step 3: Solve for N○ Expand

We are given that the probability of getting a head is 916. Equating the derived probability with the given value:

N+22(N+1)=916

Simplify the equation:

N+2N+1=98

Cross-multiply and solve for N:

8(N+2)=9(N+1)
8N+16=9N+9
16−9=9N−8N
N=7
💡 Teacher's Secret Hint

Double-check your algebraic manipulation to avoid calculation errors.

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