There are 8 boys and 5 girls in a class. If three students are selected at random, what is the proba

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25 Jul 2017 Probability

There are 8 boys and 5 girls in a class. If three students are selected at random, what is the probability that 2 girls and 1 boy are selected?

OPtion

1) 3/21
2) 7/36
3) 1/4
4) 7/23
5) 80/285
6) 40/286
7) 8/286
8) 13/40
9) 40/143
10) None of these

Solution

Total number of ways of selecting 3 students = 13C3
Total number of selecting 1 boy = 8C1
Total number of selecting 2 Girl = 5C2

Probability = (8C1 * 5C2)/13C3 = (8*5*4/2)/(13*12*11/3*2)
=> 80/286 = 40/143

Correct Option: 9 Best Solution (25)

Mukund Manohar   7 years AGO

Total number of ways of selecting 3 students = 13C3 = 286
Total number of ways of selecting 2 girls = 5C2 = 10
Total number of ways of selecting 1 boy = 8C1 = 8
Required Probability = (10*8)/286 = 80/286 = 40/143

Like? Yes (3) | No (1)  

G.S.Kantharao   7 years AGO

Probability= 8C1+5C2/13C3
=8*10/13*22=40/143. Right option is 9.👈

Like? Yes (3) | No (1)  

satish kumar   7 years AGO

total=13c3=286
2girl=5c2=10,1boy=8c1=8
p=10*8/286=40/143

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vibha   7 years AGO

((5!/21*31)*(8!/7!*1!))/13!/10!*3!

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Ravi   7 years AGO

5c2×8c1/13c3=40/143

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SAYAN CHATTERJEE   7 years AGO

Probability= (5C2X8C1/13C3)
= 40/143.
Option 9.

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Gokul Krishnan   7 years AGO

1 boy out of 8=8c1
2 girls out of 5=5c2
total 3 be chosen out of (8+5)=13c3
ans:40/143

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VARUN   7 years AGO

prob=(5c2 * 8c1)/(13c3)=40/143

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Ramkumar saini   7 years AGO

n(s)= 3 student out of 13
so 13*12*11/3*2*1=286
now (5*4/2*1)*8=80
80/286= 40/143 ans

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Nabendu   7 years AGO

(8C1*5C2)/13C3=(8*((5*4)/2))/((13*12*11)/(3*2))=40/143

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ManikumarSoruvu   7 years AGO

Number of ways of selecting 2 girls and a boy out of 8 boys and 5 girls=8c1*5c2=80
Similarly ,The number of ways of selecting 3 students out of 13 students=13c3=286
Therefore ,Required Probality=80/286=40/143

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Deepthi Sarvani   7 years AGO

8C1*5C2/13C3
8*10*3*2/(13*12*11)
=40/143
optn 9

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Sasirekha   7 years AGO

from 5 boys 1 are selected and from 8 girls 2 are selected so probability is 40/143

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SrijaMohan   7 years AGO

8 boys+5girls=13 students
3 are selected so 13C3=total (i.e) 286
from 8 boys, 1 select and 5 girls 2 select
so 5C2*8C1=10*8=80
80/286=40/143

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Priyanka   7 years AGO

8Boys, 5girls
Total=13
3 are selected at random so probability is 13c2=286
we have to select 2girls and 1 boy then the probability is 5c2*8c1=80
then 80/286=40/143

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Shazia Perween   7 years AGO

n(S)=13C3 = 286
n(E)= 8C1*5C2 = 80
therefore,,
P(E) = 80/286 = 40/143

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jyothi chhabria   7 years AGO

total probability is (8+5)c3=13c3
for selecting 2 girls and 1 boy from 8 boys and 5 gals is 5c2*8c1
I.e probality is 5c2*8c1/13c3=40/143

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DEBAYAN BASU   7 years AGO

prob of selecting 1 boy is 8C1=8
prob of selecting 2 girls 5C2=10
prob of selecting 3 students is 13C3=286
thus ans should be = (8*10)/286
= 40/143

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NABA KISHORE PADHY   7 years AGO

required probability is c(8,1)*c(5,2)/c(13,3)=40/143

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santhiya   7 years AGO

since there are 8 boys and 5 girls they can be selected in 8c1 and 5c2 from 13c2 ways

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jahanvi   7 years AGO

Number of ways of choosing two girls an done Boy is 5c2 ×8c1
number of ways of choosing any three students from the class is 13c3
probability =5c3× 8c1/13c3

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JISHA K   7 years AGO

no.of favourable possibility =5c2*8c1
total no of possibilities =13c3
probability=((5c2*8c1)/(13c2))=40/143

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Mahalakshmi   7 years AGO

13C3=236
5C2*8C1=80
Probability=80/236=40/143

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Arit   7 years AGO

P(selecting 2 girls out of 5)= 5C2
P(selecting 1 boy out of 8)= 8C2

P(selecting 3 students out of 8+5 students) = 13C3

:. P(selecting 1 boy and 2 girls out of 13 students ) = (8C1 * 5C2)/13C3 = 40/143

hence,option 9

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Ganesh Kavthale   7 years AGO

no of total ways: 286
no of ways of incident: 80
so probability :40/143

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