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Three students, Mark, Peter, and Wanda, are all working on

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Three students, Mark, Peter, and Wanda, are all working on

by BTGmoderatorDC » Wed Aug 14, 2019 9:34 pm

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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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by Ian Stewart » Thu Aug 15, 2019 5:28 am
The probability they all answer the question incorrectly is (3/4)(3/5)(5/8) = 9/32, so the probability at least one answers correctly is 1 - (9/32) = 23/32.
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by deloitte247 » Sat Aug 17, 2019 2:10 pm
Probability that
Mark is successful = 1/4
Peter is successful = 2/5
Wanda is successful = 3/8
So, the probability that
$$Mark\ is\ not\ successful\ =\frac{4}{4}-\frac{1}{4}=\frac{\left(4-1\right)}{4}=\frac{3}{4}$$
$$Peter\ is\ not\ successful\ =\frac{5}{5}-\frac{2}{5}=\frac{\left(5-2\right)}{5}=\frac{3}{5}$$
$$Wanda\ is\ not\ successful\ =\frac{8}{8}-\frac{3}{8}=\frac{\left(8-3\right)}{8}=\frac{5}{8}$$
Probability that no student will get the answer correctly =
$$=\frac{3}{4}\cdot\frac{3}{5}\cdot\frac{5}{8}=\frac{\left(3\cdot3\right)}{4\cdot8}=\frac{9}{32}$$
Probability that at least one of the students was successful =
$$=\frac{1}{1}-\frac{9}{32}=\frac{\left(32-9\right)}{32}=\frac{23}{32}$$
Answer = option C
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by Scott@TargetTestPrep » Fri Aug 23, 2019 10:38 am
BTGmoderatorDC wrote: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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by Brent@GMATPrepNow » Fri Aug 23, 2019 11:20 am
BTGmoderatorDC wrote: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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