An unbiased coin is tossed 5 times. Suppose that a variable X is assigned the value of k when k consecutive heads are obtained for k = 3, 4, 5, otherwise X takes the value -1. Then the expected value of X, is :

Solution

Maths sir · 3 steps

Read X as the longest run of heads, then count the 32 sequences carefully. The run of 3 has a free slot.

1. What X means. There are 2⁵ = 32 equally likely sequences. X is the length of the longest block of consecutive heads when that length is 3, 4 or 5. For every other sequence, X is -1.
2. Count each case. A run of exactly 3 needs a T (or the end of the sequence) on both sides of the HHH block. The fifth toss is free when the block touches an end, so it can be H or T.

   Value of X	Sequences	Count	Probability
   3	HHHT_, _THHH, THHHT (_ is H or T)	2 + 2 + 1 = 5	5/32
   4	HHHHT, THHHH	2	2/32
   5	HHHHH	1	1/32
   -1	all the rest	32 - 8 = 24	24/32

   E(X) = (3(5) + 4(2) + 5(1) + (-1)(24))/32 = (15 + 8 + 5 - 24)/32 = 4/32

3. So. E(X) = 1/8, which is option (c).

The idea to keep. For run-length problems, place the block, put a T next to it as a wall, and let every leftover toss be free. Always check that your counts add up to 2ⁿ.

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