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30 Jun 2022, 09:51
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E
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Difficulty:
75%
(hard)
Question Stats:
57% (02:31) correct
43%
(02:43)
wrong
based on 115
sessions
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A pair of fair dice is thrown independently three times. The probability of getting a total of exactly 9 twice is
(a) 1/729
(b) 8/9
(c) 1/9
(d) 8/729
(e) 8/243
(a) 1/729
(b) 8/9
(c) 1/9
(d) 8/729
(e) 8/243
30 Jun 2022, 10:18
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Iotaa
Probability of getting a total of 9 in a single throw
4/36 = 1/9
Probability of getting a total of 9 exactly in double-throw = 3C2 * (1/9)ˆ2 * (8/9) = 8/243
Updated on: 06 May 2025, 23:51
Originally posted by BrushMyQuant on 09 Oct 2022, 09:58.
Last edited by BrushMyQuant on 06 May 2025, 23:51, edited 2 times in total.
Last edited by BrushMyQuant on 06 May 2025, 23:51, edited 2 times in total.
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A pair of fair dice is thrown independently three times
As we are rolling a pair of dice => Number of cases = \(6^2\) = 36
Find The probability of getting a total of exactly 9 twice is
Sum will be 9 can happen in 4 ways
(3,6) and (6,3)
(4,5) and (5,4)
=> P(Total is 9) = \(\frac{4}{36}\) = \(\frac{1}{9}\)
=> P(Total is NOT 9) = 1 - \(\frac{1}{9}\) = \(\frac{8}{9}\)
Probability that Total is 9 is 2 out of the three rolls Can be found by picking two cases out of 3 in 3C2 ways
=> \(\frac{3!}{2!*(3-2)!}\) = 3 ways
=> Probability of getting a total of exactly 9 twice = 3 * \(\frac{1}{9}\) * \(\frac{1}{9}\) * \(\frac{8}{9}\) = \(\frac{8}{243}\)
[ 3 for picking 2 out of the three places, \(\frac{1}{9}\) for getting sum 9 twice, \(\frac{8}{9}\) for NOT getting a sum of 9 ]
So, Answer will be E
Hope it helps!
Watch the following video to learn How to Solve Dice Rolling Probability Problems
As we are rolling a pair of dice => Number of cases = \(6^2\) = 36
Find The probability of getting a total of exactly 9 twice is
Sum will be 9 can happen in 4 ways
(3,6) and (6,3)
(4,5) and (5,4)
=> P(Total is 9) = \(\frac{4}{36}\) = \(\frac{1}{9}\)
=> P(Total is NOT 9) = 1 - \(\frac{1}{9}\) = \(\frac{8}{9}\)
Probability that Total is 9 is 2 out of the three rolls Can be found by picking two cases out of 3 in 3C2 ways
=> \(\frac{3!}{2!*(3-2)!}\) = 3 ways
=> Probability of getting a total of exactly 9 twice = 3 * \(\frac{1}{9}\) * \(\frac{1}{9}\) * \(\frac{8}{9}\) = \(\frac{8}{243}\)
[ 3 for picking 2 out of the three places, \(\frac{1}{9}\) for getting sum 9 twice, \(\frac{8}{9}\) for NOT getting a sum of 9 ]
So, Answer will be E
Hope it helps!
Watch the following video to learn How to Solve Dice Rolling Probability Problems
