Multiple choice

A bag contain $(n + 1)$ coins. It is known that one of these has a head on both sides whereas the other coins are normal. One of these coins is selected at random & tossed. If the probability that the toss results in head is $\dfrac{7}{12}$, then the value of $n$ is

  1. $5$
  2. $6$
  3. $4$
  4. $3$
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A Correct answer
Explanation

Let the number of double-headed coins be 1 and normal coins be n. Total coins = n + 1. Probability of picking a double-headed coin is 1/(n+1), and a normal coin is n/(n+1). Probability of head = (1 * 1/(n+1)) + (0.5 * n/(n+1)) = 7/12. Solving (1 + 0.5n) / (n+1) = 7/12 gives 12 + 6n = 7n + 7, so n = 5.

AI explanation

The total probability of getting a head is the sum of the probabilities of drawing the two-headed coin and drawing a normal coin. The probability of drawing the two-headed coin is 1/(n+1) and it always yields a head, while the probability of drawing a normal coin is n/(n+1) and yields a head with a probability of 1/2. The combined probability is 1/(n+1) + n/2(n+1) = (n+2)/2(n+1). Setting this equal to 7/12 gives 12n + 24 = 14n + 14, so 2n = 10. Solving for n yields 5.

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