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

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

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New post 29 Mar 2017, 04:51
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A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

65% (01:00) correct 35% (00:41) wrong based on 113 sessions

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

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New post 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

Answer is C. 1/72
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Re: What is the probability of rolling three fair dice and having all thre [#permalink]

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New post 31 Mar 2017, 12:43
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.

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

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New post 31 Mar 2017, 13:17
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

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

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