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A poll is given, showing 80% are in favor of a new building project.If 6 people are chosen at random, what is the probability that exactly 2 of them favor the new building project?

User Murielle
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Given:

A poll is given, showing 80% are in favor of a new building project.

That is, the probability of success is p=0.8

Sample size, n=6

The number of times for a specific outcome within n trials is x=2.

To find the probability that exactly 2 of them favor the new building project:

Using the binomial probability,


P\mleft(x\mright)=^nC_x\cdot p^x\cdot\mleft(1-p\mright)^(n-x)

Substituting the given values, we get,


\begin{gathered} P(2)=^6C_2(0.8)^2(1-0.8)^(6-2) \\ =(6!)/((6-2)!2!)(0.64)(0.2)^4 \\ =15*0.64*0.0016 \\ =0.01536 \end{gathered}

Thus, the probability that exactly 2 of them favor the new building project is 0.01536.

User Dhaval Chheda
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