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:
A. $$\frac{{21}}{{46}}$$
B. $$\frac{{25}}{{117}}$$
C. $$\frac{{1}}{{50}}$$
D. $$\frac{{3}}{{25}}$$
Answer: Option A
Solution(By Examveda Team)
Let S be the sample space and E be the event of selecting 1 girl and 2 boysThen, 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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Comments ( 2 )
Related Questions on Probability
A. $$\frac{{1}}{{2}}$$
B. $$\frac{{2}}{{5}}$$
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What is the probability of getting a sum 9 from two throws of a dice?
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I think it should equal 7/46 because to calculate it's a conditional probability which is calculated that way = first probability * 2nd * 3rd
and the first one here equals 10/25
2nd equals 15/24
3rd equals 14/23
Can u explain in better way