Internet anywhere

In school, 60% of pupils have access to the internet at home. A group of 8 students is chosen at random. Find the probability that
a) exactly 5 have access to the internet.
b) At least 6 students have access to the internet

Correct result:

a =  0.279
b =  0.315

Solution:

$C_{{ 5}}(8) = \dbinom{ 8}{ 5} = \dfrac{ 8! }{ 5!(8-5)!} = \dfrac{ 8 \cdot 7 \cdot 6 } { 3 \cdot 2 \cdot 1 } = 56 \ \\ p=60/100=\dfrac{ 3 }{ 5 }=0.6 \ \\ q=1-p=1-0.6=\dfrac{ 2 }{ 5 }=0.4 \ \\ \ \\ a={ { 5 } \choose 8 } \cdot \ p^5 \cdot \ q^{ 8-5 }={ { 5 } \choose 8 } \cdot \ 0.6^5 \cdot \ 0.4^{ 8-5 }=0.279$

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