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If Jesse flips a coin seven times in a row, what is the probability that the result will be heads at least five times?

A. 21/128 B. 29/128 C. 35/128 D. 1/16 E. 1/4

Hi, since the Q asks us atleast 5 times, we will have to find 5 times, 6 times and 7 times.. 1) 5 times.. there are 7C5 ways that the heads can be placed in 7 times.. the prob of heads or tails is 1/2.. so prob in each way of head=(1/2)^5*(1/2)^2.. total ways=7C5*(1/2)^5*(1/2)^2= 21*(1/2)^7.. 2)6 times 7C6*(1/2)^6*(1/2)^1=7*(1/2)^7. 3)7 times 7C7*(1/2)^7*(1/2)^0=1*(1/2)^7...

total ways.. (21+7+1)*(1/2)^7=29/128 B
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If Jesse flips a coin seven times in a row, what is the probability that the result will be heads at least five times?

A. 21/128 B. 29/128 C. 35/128 D. 1/16 E. 1/4

Hi, I have explained the way to get this formula, but you can use this ai almost all 'atleast times' probability.. This is also called binomial prob formula.. prob of doing a thing x times out of total n ways= \(nCx*p^x\) *(1 − p)^n−x , where n denotes the number of trials and p denotes the success probability.
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Re: If Jesse flips a coin seven times in a row, what is the probability th [#permalink]

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09 Oct 2016, 10:20

shasadou wrote:

If Jesse flips a coin seven times in a row, what is the probability that the result will be heads at least five times?

A. 21/128 B. 29/128 C. 35/128 D. 1/16 E. 1/4

what a tricky damn one... since order doesn't matter...i started with the number of ways we can get 5, 6, or even 7 heads... 5 heads = 7C5 = 21 6 heads = 7C6 = 7 7 heads = 7C7 = 1

total: 21+7+1 = 29. this number, multiply by (1/2)^7. 1/2^7 = 1/128. 29/128 is the answer.

Re: If Jesse flips a coin seven times in a row, what is the probability th [#permalink]

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26 Oct 2017, 05:35

Probability = favorable outcomes / total outcomes

Favourable outcomes = no of times we get heads AT LEAST 5 times = no of times we get heads 5 times + no of times we get heads 6 times + no of times we get heads 7 times = 7C5 + 7C6 + 7C7 = 29