Alice, Benjamin, and Carol each try independently to win a c

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BTGmoderatorDC: Moderator; Posts: 7187; Joined: Thu Sep 07, 2017 4:43 pmFollowed by: 23 members

by BTGmoderatorDC » Thu Jan 03, 2019 2:36 am

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Alice, Benjamin, and Carol each try independently to win a carnival game. If their individual probabilities for success are 1/5, 3/8, and 2/7, respectively, what is the probability that exactly two of the three players will win but one will lose?

A. 3/140
B. 1/28
C. 3/56
D. 3/35
E. 7/40

OA E

Source: Veritas Prep

Source: Beat The GMAT — Problem Solving | Thu Jan 03, 2019 2:36 am

fskilnik@GMATH: GMAT Instructor; Posts: 1449; Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members; Thanked: 59 times

by fskilnik@GMATH » Thu Jan 03, 2019 3:05 am

BTGmoderatorDC wrote:Alice, Benjamin, and Carol each try independently to win a carnival game. If their individual probabilities for success are 1/5, 3/8, and 2/7, respectively, what is the probability that exactly two of the three players will win but one will lose?

A. 3/140
B. 1/28
C. 3/56
D. 3/35
E. 7/40
Source: Veritas Prep

$$? = P\left( {A \cap B \cap \overline C } \right) + P\left( {A \cap C \cap \overline B } \right) + P\left( {B \cap C \cap \overline A } \right)$$
$$\left( * \right)\,\,\left\{ \matrix{
\,P\left( {A \cap B \cap \overline C } \right) = {1 \over 5} \cdot {3 \over 8} \cdot {5 \over 7} = {{15} \over {5 \cdot 8 \cdot 7}} \hfill \cr
\,P\left( {A \cap C \cap \overline B } \right) = {1 \over 5} \cdot {2 \over 7} \cdot {5 \over 8} = {{10} \over {5 \cdot 7 \cdot 8}} \hfill \cr
\,P\left( {B \cap C \cap \overline A } \right) = {3 \over 8} \cdot {2 \over 7} \cdot {4 \over 5} = {{24} \over {8 \cdot 7 \cdot 5}} \hfill \cr} \right.$$
$$? = {{49} \over {5 \cdot 7 \cdot 8}} = {7 \over {40}}$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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Portuguese-speakers :: https://www.gmath.com.br

swerve: Legendary Member; Posts: 2535; Joined: Sun Oct 29, 2017 2:04 pmFollowed by: 6 members

by swerve » Thu Jan 03, 2019 8:56 am

The probability that exactly two of the three players will win = probability of (A will win, B will win, C will lose + A will win, B will lose, C will win + A will lose, B will win, C will win)

=> (1/5 * 3/8 * (1- 2/7)) + (1/5 * (1 - 3/8) * 2/7) + ((1 - 1/5) * 3/8 * 2/7)
=> 15/280 + 10/280 + 24/280
=> 49/280
=> 7/40

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by [email protected] » Fri Jan 04, 2019 10:26 am

Hi jain2016,

Since we have the probabilities of each person winning a game, we can 'map out' the 3 situations in which 2 of them win and 1 of them loses:

A wins, B wins, C loses = (1/5)(3/8)(5/7) = 15/240
A wins, B loses, C wins = (1/5)(5/8)(2/7) = 10/240
A loses, B wins, C wins = (4/5)(3/8)(2/7) = 24/240

Total probability = 15/280 + 10/280 + 24/280 = 49/280 = 7/40

Final Answer: E

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8132; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

by Scott@TargetTestPrep » Fri Feb 01, 2019 5:40 pm

BTGmoderatorDC wrote:Alice, Benjamin, and Carol each try independently to win a carnival game. If their individual probabilities for success are 1/5, 3/8, and 2/7, respectively, what is the probability that exactly two of the three players will win but one will lose?

A. 3/140
B. 1/28
C. 3/56
D. 3/35
E. 7/40

OA E

Source: Veritas Prep

We must individually consider each possible outcome of having two winners and one loser.

If Alice and Benjamin win and Carol loses, we have:

1/5 x 3/8 x 5/7 = 1 x 3/8 x 1/7 = 3/56

If Alice and Carol win and Benjamin loses, we have:

1/5 x 5/8 x 2/7 = 1 x 1/8 x 2/7 = 2/56

If Benjamin and Carol win and Alice loses, we have:

4/5 x 3/8 x 2/7 = 1/5 x 3/2 x 2/7 = 1/5 x 3 x 1/7 = 3/35

Therefore, the probability that two of them will win and one will lose is:

3/56 + 2/56 + 3/35 = 15/280 + 10/280 + 24/280 = 49/280 = 7/40

Answer: E

Scott Woodbury-Stewart
Founder and CEO
[email protected]

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