A coin is tossed "n" times.The probability of getting at least one head is greater than that of getting at least two tails by 7/128, then n=?

Asked by Vinayak Padarha | 30th Apr, 2013, 02:44: PM

Expert Answer:

A coin is tossed n times, each time the P(H) = 1/2 and P(T) = 1/2
 
Hence, P(Atleast 1 head) = 1 - P(0 heads)
 P(Atleast 1 head) = 1 - P(T)*P(T)*.....n times
 P(Atleast 1 head) = 1 - (1/2)^n
 
Also,  P(Atleast 2 tails) = 1 - P(0 tails) - P(1tail)
 P(Atleast 2 tails) = 1 - [1/2]^n - nC1 *(1/2)^n
 P(Atleast 2 tails) = 1 - [1/2]^n - n *(1/2)^n
 
Hence, as per the question
P(Atleast 1 head) - P(Atleast 2 tails) = 7/128
[ 1 - (1/2)^n] - [1 - [1/2]^n - n *(1/2)^n] = 7/128
n *(1/2)^n = 7/128
n = 7

Answered by  | 1st May, 2013, 07:31: PM

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