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# What is the probability of rolling three fair dice and having all thre

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Math Expert
Joined: 02 Sep 2009
Posts: 42558

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What is the probability of rolling three fair dice and having all thre [#permalink]

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29 Mar 2017, 04:51
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35% (medium)

Question Stats:

64% (01:00) correct 36% (00:42) wrong based on 110 sessions

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What is the probability of rolling three fair dice and having all three dice show the same odd number?

A. 71/72
B. 1/36
C. 1/72
D. 1/108
E. 1/216
[Reveal] Spoiler: OA

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Kudos [?]: 135303 [0], given: 12686

Senior Manager
Joined: 24 Apr 2016
Posts: 334

Kudos [?]: 196 [0], given: 48

Re: What is the probability of rolling three fair dice and having all thre [#permalink]

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29 Mar 2017, 11:13
Total number of possibilities = 6*6*6 = 216
Number of ways same odd number shows up in all three dice = 3 {(1,1),(3,3),(5,5)}
Probability of rolling three fair dice and having all three dice show the same odd number = 3/216

= 1/72

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Target Test Prep Representative
Affiliations: Target Test Prep
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Re: What is the probability of rolling three fair dice and having all thre [#permalink]

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31 Mar 2017, 12:43
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Bunuel wrote:
What is the probability of rolling three fair dice and having all three dice show the same odd number?

A. 71/72
B. 1/36
C. 1/72
D. 1/108
E. 1/216

The probability of the first die’s showing an odd number is 1/2. The probability of the second die’s showing the same number is 1/6. The probability of the third (or last) die’s showing the same number is also 1/6.

Thus, the overall probability is 1/2 x 1/6 x 1/6 = 1/72.

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Jeffery Miller

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e-GMAT Representative
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What is the probability of rolling three fair dice and having all thre [#permalink]

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31 Mar 2017, 13:17
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Bunuel wrote:
What is the probability of rolling three fair dice and having all three dice show the same odd number?

A. 71/72
B. 1/36
C. 1/72
D. 1/108
E. 1/216

Solution

• We need to get an odd number on each of three dice and all of them must be same.

• The odd numbers in a dice are 1,3 and 5.

o Thus, our favourable cases are (1,1,1) (3,3,3) ad (5,5,5)
• Total cases i.e the the number of ways in which we can get numbers on 3 dice $$= 6 * 6 * 6$$

o Hence Probability $$= \frac{(Favorable cases)}{(Total Cases)} = \frac{3}{(6*6*6)} = \frac{1}{72}$$
• Hence, the correct answer is Option C

Thanks,
Saquib
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Manager
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Re: What is the probability of rolling three fair dice and having all thre [#permalink]

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03 Aug 2017, 03:06
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Two ways to solve it
1st way:
No of favorable cases=3 [3 dice with odd number rolls]
Total outcomes=6*6*6=216
Probability=No of favorable cases/Total outcomes=3/216=1/72

2nd way:
The probability of the first die’s showing an odd number is 1/2. The probability of the second die’s showing the same number is 1/6. The probability of the third (or last) die’s showing the same number is also 1/6.
Thus, the overall probability is 1/2 x 1/6 x 1/6 = 1/72.
Option C.

Kudos [?]: 2 [1], given: 166

Re: What is the probability of rolling three fair dice and having all thre   [#permalink] 03 Aug 2017, 03:06
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