M4maths Previous Puzzles

26 May 2017 Probability

There are 12 boys and 9 girls in a class. If three students are selected at random, what is the probability that 1 girl and 2 boys are selected?

OPtion

1) 3/21
2) 7/36
3) 1/4
4) 7/23
5) 297/656
6) 291/665
7) 8/25
8) 297/665
9) 1/3
10) None of these

Solution

Total number of ways of selecting 3 students = 21C3
Total number of selecting 2 boys = 12 C2
Total number of selecting 1 Girl = 9C1

Probability = (12C2 * 9C1)/21C3 = ((12*11/2)*9)/(21*20*19/3*2)
= 6*11*9/7*10*19 = 297/665

Correct Option: 8 Best Solution (9)

ganesh chemanchula   8 years AGO

total number of boys and girls 12+9= 21
probability of selecting 3 from 21= 21c3=1330
probability of selecting 2 boys from 12 and 1 girl from 9=12c2.9c1=594
total probability = 594/1330=297/665

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Vinothini   8 years AGO

(12+9)C3= 21C3= 1330
9C1 * 12C2 = 9*6*11 = 594
Ans : 594 / 1330 = 297 / 665

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Amarnath Kumar Upadhyay   8 years AGO

total no of samples=21c3=1330
2 boys and 1 girl selected in 12c2*9c1 ways=594
probability is N(E)/N(S)=594/1330=297/665 ans

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Chinthalachervu.silpa   8 years AGO

12boys+ 9 girls:Total=21
3 selected at random that means 21c3
1 girl selected:9c 1
2 boys selected:12c 2
probability:(9c1×12c2)÷21c3
Ans:297/665

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kunal baghel   8 years AGO

total students = 21 then, p(selecting 3 student at random) = 21C3 = 1330
after that p(selecting 1 girld nd 2 boys) = 9C1 x 12C2 = 594

594/1330 = 297/665

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Mukund Manohar   8 years AGO

Required Probability = (12C2 * 9C1) / 21C3
= (33*9) / (7*5*19)
= 297/665

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Devendra Marghade   8 years AGO

Required probability =Selecting 1 girl and 2 boys from 9 girls and 12 boys / Selecting 3 from total 21 students
=(9C1 x 12C2) / 21C3
= 594 / 1330
= 297/665
Option 8)

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vashali   8 years AGO

1 girl can be selected out of 9 girl in 9c1 ways
2 boys can be selected out of 12 in 12c2
therefore answer willbe (9c1*12c2)/21c3

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Gokul Krishnan   8 years AGO

total students=21c3=1330
1 girl and 2 boys=594
probability =297/665

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