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24 May 2012, 01:31
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B
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:
68% (02:03) correct
32%
(02:28)
wrong
based on 1749
sessions
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Gambling with 4 dice, what is the probability of getting an even sum?
A. 3/4
B. 1/2
C. 2/3
D. 1/4
E. 1/3
A. 3/4
B. 1/2
C. 2/3
D. 1/4
E. 1/3
24 May 2012, 01:52
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cyberjadugar
Gambling with 4 dice, what is the probability of getting an even sum?
A. 3/4
B. 1/2
C. 2/3
D. 1/4
E. 1/3
Looking for a one-liner answer...
A. 3/4
B. 1/2
C. 2/3
D. 1/4
E. 1/3
Looking for a one-liner answer...
Even sum can be obtained in following cases:
EEEE - one case. Each E can take 3 values (2, 4, or 6), so total for this case is 3^4;
OOOO - one case. Each O can take 3 values (1, 3, or 5), so total for this case is 3^4;
EEOO - \(\frac{4!}{2!2!}=6\) cases (EOEO, OOEE, ...). Each E can take 3 values (2, 4, or 6) and each O can also take 3 values (1, 3, or 5), so total for this case is 6*3^2*3^2=6*3^4;
Total # of outcomes when throwing 4 dice is 6^4.
\(P=\frac{3^4+3^4+6*3^4}{6^4}=\frac{1}{2}\).
Answer: B.
Even without any math: the probability of getting an even sum when throwing 4 dice is the same as getting an even number on one die, so it must be 1/2.
Similar problem to practice: a-box-contains-100-balls-numbered-from-1-to-100-if-three-b-109279.html
Hope it helps.
SchruteDwight
Joined: 03 Sep 2018
Last visit: 30 Mar 2023
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14 Oct 2019, 06:17
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\(OOOO \implies\) 1 way (and even)
\(EOOO \implies\) 4 ways
\(EEOO \implies\) 6 ways (and even)
\(EEEO \implies\) 4 ways
\(EEEE \implies\) 1 way (and even)
Therefore
\(\frac{(1+6+1)}{(1+6+1+4+4)}\)
\(EOOO \implies\) 4 ways
\(EEOO \implies\) 6 ways (and even)
\(EEEO \implies\) 4 ways
\(EEEE \implies\) 1 way (and even)
Therefore
\(\frac{(1+6+1)}{(1+6+1+4+4)}\)
