Suppose that an unfair coin comes up heads 527 of the time T

Suppose that an unfair coin comes up heads 52.7% of the time. The coin is flipped a total of 17 times. What is the probability that you get at least 10 tails? What is the probability that you get at most 15 heads?

Solution

Suppose that an unfair coin comes up heads 52.7% of the time. The coin is flipped a total of 17 times. What is the probability that you get at least 10 tails? What is the probability that you get at most 15 heads?

Solution :

P(head) = 0.527

P(tail) = 1 - 0.527 = 0.473

Flipped 17 times.....

P(atleast 10 tails) :

= P(10T , 7H) + P(11T,6H) + P(12T , 5H) + P(13T , 4H) + P(14T , 3h) , P(15T , 2H) + P(16T , 1H) + P(17T , 0H)

= nCr(17,10)*(0.473)^10*(0.527)^7 + nCr(17,11)*(0.473)^11*(0.527)^6 + nCr(17,12)*(0.473)^12*(0.527)^5 + nCr(17,13)*(0.473)^13*(0.527)^4 + nCr(17,14)*(0.473)^14*(0.527)^3 + nCr(17,15)*(0.473)^15*(0.527)^2 + nCr(17,16)*(0.473)^16*(0.527)^1 + nCr(17,17)*(0.473)^17*(0.527)^0

= 0.239154259754634282728 ---> FIRST ANSWER

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What is the probability that you get at most 15 heads?

P(atmost 15 heads) = 1 - P(16H,1T) - P(17H,0T)

1 - nCr(17,16)*(0.527)^16*(0.473)^1 - nCr(17,17)*(0.527)^17*(0.473)^0

0.999696716138356896 ---> SECOND ANSWER