Question

At a local community college, 51% of students who enter the college as freshmen go on to graduate. Six freshmen are randomly selected.

a. What is the probability that none of them graduates from the local community college? (Do not round intermediate calculations. Round your final answer to 4 decimal places.

b. What is the probability that at most five will graduate from the local community college? (Do not round intermediate calculations. Round your final answer to 4 decimal places.)

c. What is the expected number that will graduate? (Round your final answer to 2 decimal places.)

Answer 1

$\mathbf{Solution}$

$\mathbf{given:}\;p=0.51\;(graduates),q=1-p=0.49,\;n=6$

$\mathbf{formula:}\;P(X=r)=_{r}^{n}\textrm{C}*p^{r}*q^{n-r}\;\;\;where,_{r}^{n}\textrm{C}=\frac{n!}{r!(n-r)!}$

${\color{Blue} \mathbf{a.}}\;none \;of\;them\;graduate\Rightarrow P(X=0)$

$P(X=0)=_{0}^{6}\textrm{C}*0.51^{0}*0.49^{6-0}$

$P(X=0)=\frac{6!}{0!(6-0)!}*0.51^{0}*0.49^{6}$

$P(X=0)=1*1*0.49^{6}$

$P(X=0)=\mathbf{{\color{Red} 0.0138}}$

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${\color{Blue} \mathbf{b.}}\;at\;most\;five\Rightarrow P(X\leq 5)$

$P(X\leq 5)=1-P(X>5)$

$P(X\leq 5)=1-P(X=6)$

$P(X\leq 5)=1-\left \{_{6}^{6}\textrm{C}*0.51^{6}*0.49^{6-6}\right \}$

$P(X\leq 5)=1-\left \{\frac{6!}{6!(6-6)!}*0.51^{6}*0.49^{0}\right \}$

$P(X\leq 5)=1-\left \{1*0.51^{6}*1\right \}$

$P(X\leq 5)=1-\left \{0.0176\right \}$

$P(X\leq 5)=\mathbf{{\color{Red} 0.9824}}$

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${\color{Blue} \mathbf{c.}}\;expected\;number\;E(X)=np=6*0.51=\mathbf{{\color{Red} 3.06}}$