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Question

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 $\frac{9}{16}$, then N is equal to:

The correct answer is
7

Probability Calculation for Coin Toss

Let $N$ be the number of fair coins. The total number of coins in the bag is $(N + 1)$.

  • Number of fair coins = $N$
  • Number of biased coins (Head on both sides) = $1$
  • Total number of coins = $N + 1$

Events and Probabilities

Let $F$ be the event of selecting a fair coin, and $B$ be the event of selecting the biased coin. Let $H$ be the event of getting a Head.

  • Probability of selecting a fair coin: $P(F) = \frac{N}{N + 1}$
  • Probability of selecting the biased coin: $P(B) = \frac{1}{N + 1}$
  • Probability of getting a Head given a fair coin: $P(H|F) = \frac{1}{2}$
  • Probability of getting a Head given the biased coin: $P(H|B) = 1$

Applying Law of Total Probability

The total probability of getting a Head, $P(H)$, is given by the law of total probability:

$P(H) = P(H|F) \times P(F) + P(H|B) \times P(B)$

Solving for N

We are given that $P(H) = \frac{9}{16}$. Substituting the probabilities:

$\frac{9}{16} = \left(\frac{1}{2} \times \frac{N}{N + 1}\right) + \left(1 \times \frac{1}{N + 1}\right)$

Simplify the equation:

$\frac{9}{16} = \frac{N}{2(N + 1)} + \frac{1}{N + 1}$

Combine the terms on the right side by finding a common denominator:

$\frac{9}{16} = \frac{N}{2(N + 1)} + \frac{2}{2(N + 1)}$

$\frac{9}{16} = \frac{N + 2}{2(N + 1)}$

Now, cross-multiply:

$9 \times 2(N + 1) = 16 \times (N + 2)$

$18(N + 1) = 16(N + 2)$

$18N + 18 = 16N + 32$

Isolate the terms with $N$:

$18N - 16N = 32 - 18$

$2N = 14$

Solve for $N$:

$N = \frac{14}{2}$

$N = 7$

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