anonymous
  • anonymous
help with this one plz suppose there are three students and each student tosses a coin 10 times. so what is the chance that at least one of the students gets exactly 5 heads.
Probability
jamiebookeater
  • jamiebookeater
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reemii
  • reemii
\(P(\text{at least one .... happens}) = 1 - P(\text{no such thing happens}).\) In general, this makes computations easier. You must compute \(1-P(\text{none of the 3 students got exactly 5 heads})=1-(P(A))^3\) where \(A=\{\text{one student does not get exactly 5} \}\). To compute P(A), use the same trick: P(A)=1-P(it gets exactly 5 heads).
anonymous
  • anonymous
@reemii please guide me how to calculate P(A)
reemii
  • reemii
\(P(\text{a student doesn't obtain exactly 5 heads}) = 1 - P(\text{a student obtains exactly 5 heads})\) (in 10 tosses) So we will compute : \(P(\text{5 heads in 10 tosses})\). A head appears with probability 0.5. Here you are in the situation of a binomial distribution. (\(X\sim\text{Bin}(10,0.5)\) and you want to know \(P(X=k)\)). Now it's just a formula.

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reemii
  • reemii
oops, first line is P(not 5 heads) = 1 - P(exactly 5 heads).
anonymous
  • anonymous
many thanks fot the explanation :)
anonymous
  • anonymous
0.5715

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