Question
In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is:
Answer: Option A
Let S be the sample space and E be the event of selecting 1 girl and 2 boys
Then, n(S) = Number ways of selecting 3 students out of 25
$$\eqalign{
& = {}^{25}{C_3} \cr
& = \frac{{ {25 \times 24 \times 23} }}{{ {3 \times 2 \times 1} }} \cr
& = 2300 \cr
& n\left( E \right) = {^{10}{C_1}{ \times ^{15}}{C_2}} \cr
& = {10 \times \frac{{ {15 \times 14} }}{{ {2 \times 1} }}} \cr
& = 1050 \cr
& \therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} = \frac{{1050}}{{2300}} = \frac{{21}}{{46}} \cr} $$
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Let S be the sample space and E be the event of selecting 1 girl and 2 boys
Then, n(S) = Number ways of selecting 3 students out of 25
$$\eqalign{
& = {}^{25}{C_3} \cr
& = \frac{{ {25 \times 24 \times 23} }}{{ {3 \times 2 \times 1} }} \cr
& = 2300 \cr
& n\left( E \right) = {^{10}{C_1}{ \times ^{15}}{C_2}} \cr
& = {10 \times \frac{{ {15 \times 14} }}{{ {2 \times 1} }}} \cr
& = 1050 \cr
& \therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} = \frac{{1050}}{{2300}} = \frac{{21}}{{46}} \cr} $$
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