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We flip a fair coin 10 times. What is the probability that we get heads in at least 8 of the 10 flips?

 
Guest Jun 20, 2017
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 #1
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"At least eight"   means eight or more....this is a binomial probability  problem....and we have

 

 

P ( 8 heads)  =  C(10,8) * .5^8 * .5^2

P(9 heads)  = C(10,9) * .5^9 * .5

P(10 heads)  =  .5^10

 

So.....the total probability  is

 

C(10, 8) *.5^8 * .5^2  + C(10,9) * .5^9 * .5  + .5^10  = 7/128   ≈  .0547  ≈  5.47%

 

 

cool cool cool

 
CPhill  Jun 20, 2017
 #2
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The probability is: [10C8 + 10C9 + 10C10] / 2^10 =45+10+1 / 2^10 =56/1,024 =5.46875%.

 
Guest Jun 21, 2017

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