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Exactly 2 girls

Expert replies
by harsh.champ » Wed Feb 17, 2010 5:52 am
A committee of 5 students is to be chosen from 6 boys and 4 girls. Find the probability that the committee contains exactly 2 girls.



A.1/7
B.23/24
C.11/24
D.10/21
E.None of the above.
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Source: — Problem Solving |

by ajith » Wed Feb 17, 2010 7:54 am
harsh.champ wrote:A committee of 5 students is to be chosen from 6 boys and 4 girls. Find the probability that the committee contains exactly 2 girls.



A.1/7
B.23/24
C.11/24
D.10/21
E.None of the above.

No of ways of choosing 5 students from 10 = 10C5 = 10*9*8*7*6/5*4*3*2 = 2*3*7*6

No ways of choosing 2 girls from 4 = 4c2 = 6
no of ways of choosing 2 boys from 6 = 6C2 = 15

Probability = 6*5*3/2*3*7*6 = 5/14

E
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by shashank.ism » Wed Feb 17, 2010 8:35 am
ajith wrote:
harsh.champ wrote:A committee of 5 students is to be chosen from 6 boys and 4 girls. Find the probability that the committee contains exactly 2 girls.



A.1/7
B.23/24
C.11/24
D.10/21
E.None of the above.

No of ways of choosing 5 students from 10 = 10C5 = 10*9*8*7*6/5*4*3*2 = 2*3*7*6

No ways of choosing 2 girls from 4 = 4c2 = 6
no of ways of choosing 2 boys from 6 = 6C2 = 15

Probability = 6*5*3/2*3*7*6 = 5/14

E
I think you did a mistake ajith

continuing with your solution take 3 boys instead of 2 because committee is of 5 ....so 5-2=3
no of ways of choosing 3 boys from 6 = 6C3 = 20

so probability = 6x 20/ 2x3x7x6 = 10 /21 Ans D.
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by ajith » Wed Feb 17, 2010 8:57 am
shashank.ism wrote:
I think you did a mistake ajith

continuing with your solution take 3 boys instead of 2 because committee is of 5 ....so 5-2=3
no of ways of choosing 3 boys from 6 = 6C3 = 20

so probability = 6x 20/ 2x3x7x6 = 10 /21 Ans D.
Yes, I took only 2 boys instead taking 3 in the committee thanks Shashank
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by harsh.champ » Thu Feb 18, 2010 12:14 am
shashank.ism wrote:
ajith wrote:
harsh.champ wrote:A committee of 5 students is to be chosen from 6 boys and 4 girls. Find the probability that the committee contains exactly 2 girls.



A.1/7
B.23/24
C.11/24
D.10/21
E.None of the above.

No of ways of choosing 5 students from 10 = 10C5 = 10*9*8*7*6/5*4*3*2 = 2*3*7*6

No ways of choosing 2 girls from 4 = 4c2 = 6
no of ways of choosing 2 boys from 6 = 6C2 = 15

Probability = 6*5*3/2*3*7*6 = 5/14

E
I think you did a mistake ajith

continuing with your solution take 3 boys instead of 2 because committee is of 5 ....so 5-2=3
no of ways of choosing 3 boys from 6 = 6C3 = 20

so probability = 6x 20/ 2x3x7x6 = 10 /21 Ans D.

Hey shashank,
It does happen that many-a-time when we are doing the question quickly our mind registers a different value[for example over here 3 instead of 2] especially if their values are complimentary[over here 3 = 5 - 2].Thanks for pointing it out.
No ways of choosing 2 girls from 4 = 4c2 = 6
no of ways of choosing 2 boys from 6 = 6C2 = 15
It is thus very important to keep these mistakes at bay in the actual exam.
It takes time and effort to explain, so if my comment helped you please press Thanks button :)



Just because something is hard doesn't mean you shouldn't try,it means you should just try harder.

"Keep Walking" - Johnny Walker :P
Join the discussion