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Three students, Mark, Peter, and Wanda, are all working on the same math problem. If their individual probabilities of

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Three students, Mark, Peter, and Wanda, are all working on the same math problem. If their individual probabilities of

by BTGmoderatorDC » Thu Jan 16, 2020 6:24 pm
Three students, Mark, Peter, and Wanda, are all working on the same math problem. If their individual probabilities of success are 1/4, 2/5, and 3/8, respectively, then what is the probability that at least one of the students will get the problem correct?

A. 3/80
B. 9/32
C. 23/32
D. 77/80
E. 39/40



OA C

Source: Princeton Review
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Re: Three students, Mark, Peter, and Wanda, are all working on the same math problem. If their individual probabilities

by Jay@ManhattanReview » Thu Jan 16, 2020 8:24 pm
The problem will be solved in three conditions:

1. Anyone among Mark, Peter, and Wanda solves the problem and the other two don't
2. Any two among Mark, Peter, and Wanda solve the problem and the other one doesn't
3. All three Mark, Peter, and Wanda solve the problem

So, the problem will remain unsolved only when all three are not able to solve it. So, a better approach to get to the question is to find out the probability that none could solve the problem and then deduct it from 1.

• Probability of Mark not able to solve the problem = 1 – 1/4 = 3/4;
• Probability of Peter not able to solve the problem = 1 – 2/5 = 3/5;
• Probability of Wanda not able to solve the problem = 1 – 3/8 = 5/8

Thus, the probability that none can solve the problem = (3/4)*(3/5)*(5/8) = 9/32

Thus, the probability that at least one can solve the problem = 1 – 9/32 = 23/32

The correct answer: C

Hope this helps!

-Jay
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Re: Three students, Mark, Peter, and Wanda, are all working on the same math problem. If their individual probabilities

by swerve » Sat Jan 18, 2020 2:55 pm
BTGmoderatorDC wrote: ↑
Thu Jan 16, 2020 6:24 pm
 Three students, Mark, Peter, and Wanda, are all working on the same math problem. If their individual probabilities of success are 1/4, 2/5, and 3/8, respectively, then what is the probability that at least one of the students will get the problem correct?

A. 3/80
B. 9/32
C. 23/32
D. 77/80
E. 39/40



OA C

Source: Princeton Review
CASE 1
Mark right \(1/4\); wrong \(3/4\)
Peter right \(2/5\); wrong \(3/5\)
Wanda right \(3/8\); wrong \(5/8\)

So, we get following cases:
\(1/4\cdot 3/5\cdot 5/8 + 2/5\cdot 3/4 \cdot 5/8 + 3/4\cdot 3/5\cdot 3/8 + 1/4\cdot 2/5\cdot 5/8 + 1/4\cdot 3/8\cdot 3/5 + 2/5\cdot 3/8\cdot 3/4 + 1/4\cdot 2/5\cdot 3/8\); \(115/160\); \(23/32\). Therefore, __C__

CASE 2
Else all wrong cases, \(3/4\cdot 3/5\cdot 5/8 = 9/32\)
1- wrong cases ; \(1-9/32; 23/32\). Therefore, __C__
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Re: Three students, Mark, Peter, and Wanda, are all working on the same math problem. If their individual probabilities

by Scott@TargetTestPrep » Thu Jan 23, 2020 11:09 am
BTGmoderatorDC wrote: ↑
Thu Jan 16, 2020 6:24 pm
 Three students, Mark, Peter, and Wanda, are all working on the same math problem. If their individual probabilities of success are 1/4, 2/5, and 3/8, respectively, then what is the probability that at least one of the students will get the problem correct?

A. 3/80
B. 9/32
C. 23/32
D. 77/80
E. 39/40



OA C

Source: Princeton Review
We can use the formula:

1 - P(none get the problem correct) = P(at least one gets the problem correct)

1 - P(none get the problem correct) = 1 - 3/4 x 3/5 x 5/8 = 1 - 9/32 = 23/32

Answer: C

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Re: Three students, Mark, Peter, and Wanda, are all working on the same math problem. If their individual probabilities

by Brent@GMATPrepNow » Fri Jan 24, 2020 7:49 am
BTGmoderatorDC wrote: ↑
Thu Jan 16, 2020 6:24 pm
 Three students, Mark, Peter, and Wanda, are all working on the same math problem. If their individual probabilities of success are 1/4, 2/5, and 3/8, respectively, then what is the probability that at least one of the students will get the problem correct?

A. 3/80
B. 9/32
C. 23/32
D. 77/80
E. 39/40



OA C

Source: Princeton Review
Aside: If P(Mark is correct) = 1/4, then P(Mark is INcorrect) = 3/4
If P(Peter is correct) = 2/5, then P(Peter is INcorrect) = 3/5
If P(Wanda is correct) = 3/8, then P(Wanda is INcorrect) = 5/8
---------------------------

When it comes to probability questions involving at least, it's best to try using the complement.
That is, P(Event A happening) = 1 - P(Event A not happening)

So, we can write: P(at least one student is correct) = 1 - P(NO students are correct)

P(NO students are correct) = P(Mark is incorrect AND Peter is incorrect AND Wanda is incorrect)
= P(Mark is incorrect) x P(Peter is incorrect) x P(Wanda is incorrect)
= 3/4 x 3/5 x 5/8
= 9/32

So, P(at least one student is correct) = 1 - 9/32
= 23/32

Answer: C

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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