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sonusaini1
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

Probability that no student will get the answer correct will be 3/4*3/5*5/8 = 9/32

Probability that at least one student get the answer correct will be 1- 9/32= 23/32

Choice C is correct.

Thanks & Regards

Posted from my mobile device

sonusaini1 see highlighted part ; wont 1- all wrong answer give P that all will be correct? instead of at least one of them is correct?
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Bunuel
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

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* 3/5*5/8 + 2/5*3/4*5/8+ 3/4*3/5*3/8+1/4*2/5*5/8+1/4*3/8*3/5+2/5*3/8*3/4+1/4*2/5*3/8
; 105/160 ; 21 /32
I am not sure why answer option is missing ; probably I am doing some calculation mistake ? Bunuel is given answer correct in the option ?

I think calculating none of the case probability and then subtracting it from 1 will give the answer faster.
Answer set will have 1, 2 or all three correct.
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Bunuel
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


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* 3/5*5/8 + 2/5*3/4*5/8+ 3/4*3/5*3/8+1/4*2/5*5/8+1/4*3/8*3/5+2/5*3/8*3/4+1/4*2/5*3/8
; 115/160 ; 23 /32
IMO C

CASE 2
else all wrong cases; 3/4*3/5*5/8 = 9/32
1- wrong cases ; 1-9/32 ; 23/32

IMO C

I think the shortcut is All probability - The condition when everyone gets wrong
-> This will give us at least one correct scenario probability.

OA: 1 - {(3/4)*(3/5)*(5/8)}
= 23/32
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Bunuel
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

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
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