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

Result

a =  0.279
b =  0.315

Solution:

$p = 60/100 = \dfrac{ 3 }{ 5 } = 0.6 \ \\ q = 1-p = 1-0.6 = \dfrac{ 2 }{ 5 } = 0.4 \ \\ \ \\ a = { { 5 } \choose 8 } \cdot \ p^5 \cdot \ q^{ 8-5 } = { { 5 } \choose 8 } \cdot \ 0.6^5 \cdot \ 0.4^{ 8-5 } \doteq 0.2787 = 0.279C_{{ 5}}(8) = \dbinom{ 8}{ 5} = \dfrac{ 8! }{ 5!(8-5)!} = \dfrac{ 8 \cdot 7 \cdot 6 } { 3 \cdot 2 \cdot 1 } = 56 \ \\ \ \\ a = { { 5 } \choose 8 } \cdot \ p^5 \cdot \ q^{ 8-5 } = { { 5 } \choose 8 } \cdot \ 0.6^5 \cdot \ 0.4^{ 8-5 } \doteq 0.2787 = 0.279$

$C_{{ 8}}(8) = \dbinom{ 8}{ 8} = \dfrac{ 8! }{ 8!(8-8)!} = \dfrac{ 1 } { 1 } = 1 \ \\ \ \\ p_{ 6 } = { { 6 } \choose 8 } \cdot \ p^6 \cdot \ q^{ 8-6 } = { { 6 } \choose 8 } \cdot \ 0.6^6 \cdot \ 0.4^{ 8-6 } \doteq 0.209 \ \\ p_{ 7 } = { { 7 } \choose 8 } \cdot \ p^7 \cdot \ q^{ 8-7 } = { { 7 } \choose 8 } \cdot \ 0.6^7 \cdot \ 0.4^{ 8-7 } \doteq 0.0896 \ \\ p_{ 8 } = { { 8 } \choose 8 } \cdot \ p^8 \cdot \ q^{ 8-8 } = 1 \cdot \ 0.6^8 \cdot \ 0.4^{ 8-8 } \doteq 0.0168 \ \\ \ \\ b = p_{ 6 }+p_{ 7 }+p_{ 8 } = 0.209+0.0896+0.0168 \doteq 0.3154 = 0.315C_{{ 7}}(8) = \dbinom{ 8}{ 7} = \dfrac{ 8! }{ 7!(8-7)!} = \dfrac{ 8 } { 1 } = 8 \ \\ \ \\ b = p_{ 6 }+p_{ 7 }+p_{ 8 } = 0.209+0.0896+0.0168 \doteq 0.3154 = 0.315C_{{ 6}}(8) = \dbinom{ 8}{ 6} = \dfrac{ 8! }{ 6!(8-6)!} = \dfrac{ 8 \cdot 7 } { 2 \cdot 1 } = 28 \ \\ \ \\ b = p_{ 6 }+p_{ 7 }+p_{ 8 } = 0.209+0.0896+0.0168 \doteq 0.3154 = 0.315C_{{ 7}}(8) = \dbinom{ 8}{ 7} = \dfrac{ 8! }{ 7!(8-7)!} = \dfrac{ 8 } { 1 } = 8 \ \\ \ \\ b = p_{ 6 }+p_{ 7 }+p_{ 8 } = 0.209+0.0896+0.0168 \doteq 0.3154 = 0.315$

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