Question

Consider independent trials of flipping fair coins (outcomes are heads or tails). Define the random variable T to be the first time that two heads come up in a row (so, for the outcome HT HT HH... we have T = 6).

(a) Compute P(T = i) for i = 1, 2, 3, 4, 5.

(b) Compute P(T = n) for n > 5.

Answer 1

Probability of Head, H = 1/2

Probability of Tail, T = 1/2

a. Since we need at least 2 trials of flipping fair coins to have two heads in a row.

For T =1 , P(T=1) = 0,

Both the trials have to be head

For T = 2, P(T=2) = HH = 1/2 * 1/2 = 1/4.

The first trial has to be a tail , followed by couple of heads

For T = 3, P(T=3) = THH = 1/2 * 1/2 * 1/2 = 1/8.

The last three trails have to be THH in that order.. The 1st trial can be either Head or Tail . So , we have 2 possibilities.. we calculate their probability and sum them.

For T = 4, P(T=4) = TTHH , HTHH = 1/2 * 1/2 * 1/2 * 1/2 + 1/2 * 1/2 * 1/2 * 1/2    = 1/16 + 1/16 = 1/8.

The last three trails have to be THH in that order.. The 1st two trails can be TT, TH, HT.(It cant be HH , because then it will be P( T=2)). Now, these probability of these 3 combinations are summed up to get the probability.

For T = 5, P(T=5) = TTTHH, THTHH, HTTHH = 1/2 * 1/2 * 1/2 * 1/2 * 1/2 + 1/2 * 1/2 * 1/2 * 1/2 * 1/2 + 1/2 * 1/2 * 1/2 * 1/2 * 1/2 = 3/32

b. P( T > 5) = 1 - P( T 5)

Now, P( T 5) = P(T = 0) +P(T = 1) +P(T = 2) +P(T = 3) +P(T = 4) +P(T = 5)

P( T 5) = 0 + 0 + 1/4 + 1/8 + 3/32 = 15/32

Therefore, P( T > 5) = 1 - P( T 5) = 1 - 15/32 = 17/32.

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