Answered

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Suppose that 45% of all babies born in a particular hospital are girls. If 7 babies born in the hospital are randomly selected, what is the probability that at most 2 of them are girls?Carry your intermediate computations to at least four decimal places, and round your answer to two decimal places.

Sagot :

Solution:

Since the sample size is fixed;

[tex]n=7[/tex]

And the probability of picking a girl is constant for each trial, the binomial distribution is appropriate;

[tex]\begin{gathered} P(X=r)=^nC_rP^r(1-P)^{n-r} \\ \\ P(X\leq2) \end{gathered}[/tex]

Thus;

[tex]\Rightarrow^7C_0(0.45)^0(0.55)^7+^7C_1(0.45)^1(0.55)^6+^7C_2(0.45)^2(0.55)^5[/tex][tex]\Rightarrow0.0152+0.0872+0.2140=0.3164[/tex]

ANSWER: 0.32

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