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## anonymous 3 years ago 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.

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1. 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).

2. anonymous

@reemii please guide me how to calculate P(A)

3. 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.

4. reemii

oops, first line is P(not 5 heads) = 1 - P(exactly 5 heads).

5. anonymous

many thanks fot the explanation :)

6. anonymous

0.5715

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