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rovshan85
what is the prob. of getting at least 2 heads in a row on three flips of a fair coin?

no answers are available, sorry

probability for HHT;

Total number of ways of arranging HHT keeping HH intact is 2!

probability = 2! * (1/2)^3

probability for HHH;

no of ways= 1

probability = (1/2)^3

hence required answer = 2! * (1/2)^3 + (1/2)^3
= 3* 1/8 = 3/8

Hope this helps...!!
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Banuel's explanation is again the simpliest and the steadiest against mistakes.
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rovshan85
What is the probability of getting at least 2 heads in a row on three flips of a fair coin?

We can assume the first two flips are heads (H) and the last flip is tails (T). Thus:

P(H-H-T) = 1/2 x 1/2 x 1/2 = ⅛

The only other way to get two heads in a row would be flipping heads on the second and third flips.

P(T-H-H) = 1/2 x 1/2 x 1/2 = ⅛

Thus, the total probability of getting two heads in a row when we flip a coin three times is 1/8 + 1/8 = 2/8.

Next, we need to determine the probability of getting heads on all three flips.

P(H-H-H) = 1/2 x 1/2 x 1/2 = 1/8.

Thus, the probability of getting at least two heads in a row is 2/8 + 1/8 = 3/8.
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total number case =2*2*2 =8
as listed below
HHH
HHT
THH

HTH
TTH
THT
TTT
HTT

first 3 case have alteast 2 consecutive heads, so probability is 3/8
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P(atleast 2 heads in a row) = P(2 heads in a row) + P(3 heads)

ways to get 2 heads in a row = HHT and THH
ways to get 3 heads in a row: HHH

So, final probability: 1/2^3*(3)
=3/8 (C)
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