A coin is biased so that the head is 3 times as likely to occur as tail. This coin is tossed until a head or three tails occur. If X denotes the number of tosses of the coin, then the mean of X is :

Solution

Maths sir · 4 steps

The game can last only 1, 2 or 3 tosses, so list those three cases and weight each by its probability.

1. Find the probabilities. Head is 3 times as likely as tail, so P(H) = 3P(T) and P(H) + P(T) = 1. That gives P(H) = 3/4 and P(T) = 1/4.
2. Possible values of X. Tossing stops at the first head or after the third tail. So X can only be 1, 2 or 3. At X = 3 the tossing stops whatever the third toss shows, so both TTH and TTT count.

   X	Sequences	P(X)	X · P(X)
   1	H	3/4 = 12/16	12/16
   2	TH	1/4 × 3/4 = 3/16	6/16
   3	TTH or TTT	1/4 × 1/4 = 1/16	3/16

3. Check. The probabilities add up correctly: 12/16 + 3/16 + 1/16 = 1.

   E(X) = 12/16 + 6/16 + 3/16 = 21/16

4. So. The mean number of tosses is 21/16, which is option (c).

The idea to keep. For a stopping game, list every way it can end. Make sure the probabilities add up to 1 before you compute Σ x P(x). If they do not, you have missed a case.

Answer from the NTA final answer key. Solution by PrepWiser.