Q35 / 75 · JEE Main 2026 (06 April Shift 1) · Mathematics
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 $\dfrac{9}{16}$, then $N$ is equal to:
- A$5$
- Correct answerB$7$
- C$8$
- D$9$
Solution
Let $E_1$ be the event of selecting a fair coin and $E_2$ be the event of selecting the two-headed coin.
$P(E_1) = \dfrac{N}{N+1}$
$P(E_2) = \dfrac{1}{N+1}$
Let $H$ be the event of getting a Head.
$P(H|E_1) = \dfrac{1}{2}$
$P(H|E_2) = 1$
Using the theorem of total probability:
$P(H) = P(E_1)P(H|E_1) + P(E_2)P(H|E_2)$
$P(H) = \left(\dfrac{N}{N+1}\right)\left(\dfrac{1}{2}\right) + \left(\dfrac{1}{N+1}\right)(1)$
$P(H) = \dfrac{N+2}{2(N+1)}$
Given $P(H) = \dfrac{9}{16}$:
$\dfrac{N+2}{2(N+1)} = \dfrac{9}{16}$
$\dfrac{N+2}{N+1} = \dfrac{9}{8}$
$8N + 16 = 9N + 9$
$N = 7$
Answer: $7$
$P(E_1) = \dfrac{N}{N+1}$
$P(E_2) = \dfrac{1}{N+1}$
Let $H$ be the event of getting a Head.
$P(H|E_1) = \dfrac{1}{2}$
$P(H|E_2) = 1$
Using the theorem of total probability:
$P(H) = P(E_1)P(H|E_1) + P(E_2)P(H|E_2)$
$P(H) = \left(\dfrac{N}{N+1}\right)\left(\dfrac{1}{2}\right) + \left(\dfrac{1}{N+1}\right)(1)$
$P(H) = \dfrac{N+2}{2(N+1)}$
Given $P(H) = \dfrac{9}{16}$:
$\dfrac{N+2}{2(N+1)} = \dfrac{9}{16}$
$\dfrac{N+2}{N+1} = \dfrac{9}{8}$
$8N + 16 = 9N + 9$
$N = 7$
Answer: $7$
Hence, the correct answer is option: 'B'.