Maths · Probability: Probability of an event, addition and multiplication theorems of probability

A bag contains \) coins. It is known that of these coins have a head on both sid

A bag contains \( (2 n+1) \) coins. It is known that \( n \) of these coins have a head on both sides, whereas the remaining \( n+1 \) coins are fair. A coin is picked up at random from the bag and tossed. If the probability that the toss results in a head is \( \frac{31}{42}, \) then \( n \) is equal to

  • A. 10
  • B. 11
  • C. 12
  • D. 13

Step-by-step solution

Let the total coins be 2n+1. n two-headed coins always give head, and n+1 fair coins give head with probability 1/2. Total probability of head = n/(2n+1)*1 + (n+1)/(2n+1)*(1/2) = (3n+1)/(2(2n+1)). Set equal to 31/42: (3n+1)/(2(2n+1)) = 31/42 => 42(3n+1)=62(2n+1) => 126n+42=124n+62 => 2n=20 => n=10.
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