Answer to Question #229507 in Statistics and Probability for Fatma Al Hashimi

Question #229507

The probability that a student is admitted to MCBS for bachelor is 0.6. If 10 students from the same school apply, find the probability that (i) at most 5 are admitted. (ii) at least 7 are rejected


1
Expert's answer
2021-08-25T18:26:50-0400

Let X=X= the number of admitted students: XBin(n,p).X\sim Bin(n, p).

Given p=0.6,n=10,q=1p=0.4.p=0.6, n=10, q=1-p=0.4.

(i)


P(X5)=P(X=0)+P(X=1)+P(X=2)P(X\leq5)=P(X=0)+P(X=1)+P(X=2)

+P(X=3)+P(X=4)+P(X=5)+P(X=3)+P(X=4)+P(X=5)

=(100)(0.6)0(0.4)100+(101)(0.6)1(0.4)101=\dbinom{10}{0}(0.6)^0(0.4)^{10-0}+\dbinom{10}{1}(0.6)^1(0.4)^{10-1}

+(102)(0.6)2(0.4)102+(103)(0.6)3(0.4)103+\dbinom{10}{2}(0.6)^2(0.4)^{10-2}+\dbinom{10}{3}(0.6)^3(0.4)^{10-3}

+(104)(0.6)4(0.4)104+(105)(0.6)5(0.4)105+\dbinom{10}{4}(0.6)^4(0.4)^{10-4}+\dbinom{10}{5}(0.6)^5(0.4)^{10-5}

0.3667\approx0.3667


(ii)


P(X3)=P(X=0)+P(X=1)+P(X=2)P(X\leq3)=P(X=0)+P(X=1)+P(X=2)

+P(X=3)=(100)(0.6)0(0.4)100+P(X=3)=\dbinom{10}{0}(0.6)^0(0.4)^{10-0}

+(101)(0.6)1(0.4)101+(102)(0.6)2(0.4)102+\dbinom{10}{1}(0.6)^1(0.4)^{10-1}+\dbinom{10}{2}(0.6)^2(0.4)^{10-2}

+(103)(0.6)3(0.4)1030.0548+\dbinom{10}{3}(0.6)^3(0.4)^{10-3}\approx0.0548


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