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Question

There are 15 boys and 10 girls in a class. If three students are selected at random, what is the probability that 1 girl and 2 boys are selected?
Options:
A .  $MF#%\dfrac{1}{40}$MF#%
B .  $MF#%\dfrac{1}{2}$MF#%
C .  $MF#%\dfrac{21}{46}$MF#%
D .  $MF#%\dfrac{7}{42}$MF#%
Answer: Option C

Answer : Option C

Explanation :

Let S be the sample space.
n(S) = Total number of ways of selecting 3 students from 25 students = 25C3
Let E = Event of selecting 1 girl and 2 boys
n(E) = Number of ways of selecting 1 girl and 2 boys
15 boys and 10 girls are there in a class.
We need to select 2 boys from 15 boys and 1 girl from 10 girls
Number of ways in which this can be done = 15C2 × 10C1
Hence n(E) = 15C2 × 10C1

$MF#% \text{P(E) = }\dfrac{\text{n(E)}}{\text{n(S)}}$MF#%$MF#% = \dfrac{15_{C_2} \times 10_{C_1}}{25_{C_3}}$MF#%

$MF#%\begin{align}&=\dfrac{\left(\dfrac{15 \times 14 }{2 \times1}\right) \times 10}{\left(\dfrac{25 \times 24 \times 23}{3 \times 2 \times1}\right)}
= \dfrac{15 \times 14 \times 10}{\left(\dfrac{25 \times 24 \times 23}{3}\right)}
= \dfrac{15 \times 14 \times 10}{25 \times 8 \times 23}
= \dfrac{3 \times 14 \times 10}{5 \times 8 \times 23}\\\\\\\\
&= \dfrac{3 \times 14 \times 2}{8 \times 23}
= \dfrac{3 \times 14 }{4 \times 23}
= \dfrac{3 \times 7}{2 \times 23}
= \dfrac{21}{46}\end{align} $MF#%



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