Group A has 2 boys and 3 girls, group B has 3 boys and...

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BTGmoderatorLU: Moderator; Posts: 2544; Joined: Sun Oct 15, 2017 1:50 pmFollowed by: 6 members

by BTGmoderatorLU » Thu Mar 15, 2018 2:52 pm

Group A has 2 boys and 3 girls, group B has 3 boys and 4 girls and group C has 2 boys and 4 girls. One student is selected from each of the group. Find the probability that one girl and two boys are among the three selected?

A. 3/4
B. 1/18
C. 29/105
D. 29/315
E. 77/272

The OA is C.

I'm really confused by this PS question. Please, can any expert assist me with it? I need your help. Thanks in advance.

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Source: Beat The GMAT — Problem Solving | Thu Mar 15, 2018 2:52 pm

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Jay@ManhattanReview: GMAT Instructor; Posts: 3008; Joined: Mon Aug 22, 2016 6:19 amFollowed by: 34 members; Location: Grand Central / New York; Thanked: 470 times

by Jay@ManhattanReview » Thu Mar 15, 2018 8:47 pm

LUANDATO wrote:Group A has 2 boys and 3 girls, group B has 3 boys and 4 girls and group C has 2 boys and 4 girls. One student is selected from each of the group. Find the probability that one girl and two boys are among the three selected?

A. 3/4
B. 1/18
C. 29/105
D. 29/315
E. 77/272

The OA is C.

I'm really confused by this PS question. Please, can any expert assist me with it? I need your help. Thanks in advance.

The three groups have

Group A: 2 boys and 3 Girls;
Group B: 3 boys and 4 Girls;
Group C: 2 boys and 4 Girls

We have to select one student from each of the group such that one girl and two boys are among the three selected.

Case 1: (GBB): Selecting one Girl from Group A, one Boy from Group B, and one Boy from Group C

Probability (GBB) = 3/5*3/7*2/6 = 3/35 ---(1)

Case 2: (BBG): Selecting one Boy from Group A, one Boy from Group B, and one Girl from Group C

Probability (BBG) = 2/5*3/7*4/6 = 4/35 ---(2)

Case 3: (BGB): Selecting one Boy from Group A, one Girl from Group B, and one Boy from Group C

Probability (BGB) = 2/5*4/7*2/6 = 8/105 ---(3)

Adding the three probabilities...

Probability that one girl and two boys are selected = 3/35 + 4/35 + 8/105 = 29/105.

The correct answer: C

Hope this helps!

-Jay
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Manhattan Review GMAT Prep

Locations: New York | Singapore | London | Dubai | and many more...

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8137; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

by Scott@TargetTestPrep » Mon Mar 19, 2018 3:36 pm

LUANDATO wrote:Group A has 2 boys and 3 girls, group B has 3 boys and 4 girls and group C has 2 boys and 4 girls. One student is selected from each of the group. Find the probability that one girl and two boys are among the three selected?

A. 3/4
B. 1/18
C. 29/105
D. 29/315
E. 77/272

We have 3 scenarios: 1) the girl is from group A, 2) the girl is from group B and 3) the girl is from group C.

1) The girl is from group A (and the 2 boys one each from group B and group C)

Probability = 3/5 x 3/7 x 2/6 = 18/210

2) The girl is from group B (and the 2 boys one each from group A and group C)

Probability = 2/5 x 4/7 x 2/6 = 16/210

3) The girl is from group C (and the 2 boys one each from group A and group B)

Probability = 2/5 x 3/7 x 4/6 = 24/210

Thus the overall probability is:

18/210 + 16/210 + 24/210 = 58/210 = 29/105

Answer: C

Scott Woodbury-Stewart
Founder and CEO
[email protected]

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BTGmoderatorRO: Moderator; Posts: 772; Joined: Wed Aug 30, 2017 6:29 pmFollowed by: 6 members

by BTGmoderatorRO » Sun Mar 25, 2018 7:25 am

Group A- 2 boys and 3 girls
Group B- 3 boys and 4 girls
Group C- 2 boys and 4 girls
$$probab\ ility\ of\ selecting\ the\ three\ students\ could\ be\ \left(GBB\right)\ or\ \left(BGB\right)\ or\ \left(BBG\right)$$
$$Therefore,\ \Pr\left(GBB\right)=\frac{3}{5}\cdot\frac{3}{7}\cdot\frac{2}{6}=\frac{3}{35}$$
$$Therefore,\ \Pr\left(BGB\right)=\frac{2}{5}\cdot\frac{4}{7}\cdot\frac{2}{6}=\frac{8}{105}$$
$$Therefore,\ \Pr\left(BGB\right)=\frac{2}{5}\cdot\frac{3}{7}\cdot\frac{4}{6}=\frac{4}{35}$$
$$Therefore,\ \Pr\left(GBB\right)+\Pr\left(BGB\right)+\Pr\left(BBG\right)=\Pr\left[\frac{3}{35}+\frac{8}{105}+\frac{4}{35}\right]=\frac{\left(9+8+12\right)}{105}=\frac{29}{105}$$
The correct answer is C

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