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Updated on: 06 Oct 2019, 20:56
Originally posted by almostfamous on 12 Feb 2011, 05:45.
Last edited by Bunuel on 06 Oct 2019, 20:56, edited 2 times in total.
Last edited by Bunuel on 06 Oct 2019, 20:56, edited 2 times in total.
Renamed the topic and edited the question.
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
72% (02:46) correct
28%
(02:55)
wrong
based on 1000
sessions
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Time
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Crowan throws 3 dice and records the product of the numbers appearing at the top of each die as the result of the attempt. What is the probability that the result of any attempt is an odd integer divisible by 25?
A. 7/216
B. 5/91
C. 13/88
D. 1/5
E. 3/8
A. 7/216
B. 5/91
C. 13/88
D. 1/5
E. 3/8
Could anyone enlighten me on this tricky probability problem:
"If you throw 3 fair dice and take the product of the three numbers on the top of each die that result, what is the probability that you get an even number divisible by 25 at any attempt?"
My logic is that:
1) You need at least two fives in order to get 25. So: 1/6 x 1/6 x something
2) For the "something" above, you have three options: 2, 4, or 6, because the question asks for an even number. So: 1/6 x 1/6 x 1/2 (i.e. 3 favorable outcomes out of six) = 1/72
(Source: Veritas Prep, Combinatorics p.102)
However, the OA is 7/216, and I cannot fathom what my logic above is missing out. If anyone could lend a hand to solve this problem, I would appreciate.
Thanks!
"If you throw 3 fair dice and take the product of the three numbers on the top of each die that result, what is the probability that you get an even number divisible by 25 at any attempt?"
My logic is that:
1) You need at least two fives in order to get 25. So: 1/6 x 1/6 x something
2) For the "something" above, you have three options: 2, 4, or 6, because the question asks for an even number. So: 1/6 x 1/6 x 1/2 (i.e. 3 favorable outcomes out of six) = 1/72
(Source: Veritas Prep, Combinatorics p.102)
However, the OA is 7/216, and I cannot fathom what my logic above is missing out. If anyone could lend a hand to solve this problem, I would appreciate.
Thanks!
17 Feb 2011, 14:23
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almostfamous
Crowan throws 3 dice and records the product of the numbers appearing at the top of each die as the result of the attempt. What is the probability that the result of any attempt is an ODD integer divisible by 25?
A. 7/216
B. 5/91
C. 13/88
D. 1/5
E. 3/8
In this case as 555 can occur only one way, then in nominator we'll have 3+3+1=7:
551 - 515 - 155;
553 - 535 - 355;
555.
\(P(55O)=\frac{7}{216}\) (if you want to calculate as in previous case then: \(P(55O)=P(551)+P(553)+P(555)=\frac{3!}{2!}*\frac{1}{6}*\frac{1}{6}*\frac{1}{6}+\frac{3!}{2!}*\frac{1}{6}*\frac{1}{6}*\frac{1}{6}+\frac{1}{6}*\frac{1}{6}*\frac{1}{6}=\frac{7}{216}\))
Note that above 16 scenarios describe all cases of having at least two 5 on 3 dice: 7 cases give odd sum and 9 give even sum.
Answer: A.
If you throw 3 fair dice and take the product of the three numbers on the top of each die that result, what is the probability that you get an EVEN number divisible by 25 at any attempt?
\(P(55E)=\frac{3!}{2!}*\frac{1}{6}*\frac{1}{6}*\frac{3}{6}=\frac{9}{216}\), we multiply by \(\frac{3!}{2!}\) or simply by 3 because 5-5-Even can occur in 3 different ways: 5-5-Even, 5-Even-5, or Even-5-5 (for example: 552, 525, or 255).
Check all 9 scenarios:
552 - 525 - 255;
554 - 545 - 455;
556 - 565 - 655.
Hope it helps.
12 Jun 2018, 11:12
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almostfamous
The total number of ways to throw 3 dice is 6 x 6 x 6 = 216
The only way to get a product of the numbers appearing on the top of 3 dice to be an odd number that is divisible by 25 is if two of the numbers are both 5 and the third number is an odd number. Thus, we have:
{1, 5, 5}, {5, 1, 5}, {5, 5, 1}, {3, 5, 5}, {5, 3, 5}, {5, 5, 3} and {5, 5, 5}
Thus the probability is 7/216.
Answer: A
