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17 Mar 2020, 02:26
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D
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:
62% (02:11) correct
38%
(02:31)
wrong
based on 218
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A pair of fair dice are rolled together, till a sum of either 5 or 7 is obtained. The probability that the sum 5 happens before sum 7 is
A. 1/10
B. 1/5
C. 3/10
D. 2/5
E. 3/5
Are You Up For the Challenge: 700 Level Questions
A. 1/10
B. 1/5
C. 3/10
D. 2/5
E. 3/5
Are You Up For the Challenge: 700 Level Questions
17 Mar 2020, 03:02
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Bunuel
Total Outcome for a roll of pair of dice = 6*6 = 36
The possible outcomes of 5 = {1, 4} {2, 3} {3, 2} {4, 1} = 4 cases
The possible outcomes of 7 = {1, 6} {2, 5} {3, 4} {4, 3} {5, 2} {6, 1} = 6 cases
Sum either 5 or 7 cases = 4+6 = 10
Probability of getting sum 5 before 7 = (4/36) / (10/36) = 2/5
Answer: Option D
General Discussion
17 Mar 2020, 03:27
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Probability of getting 5 = \(\frac{4}{36}\) = \(\frac{1}{9}\)
Probability of getting 7 = \(\frac{6}{36}\) = \(\frac{1}{6}\)
We don't want want sum 7 occurs before sum 5.
Case 1. We get sum 5 on the first turn.
P(x) =\( \frac{1}{9}\)
Case 2- We get sum 5 on the second turn.
\(P(x)= (1-\frac{1}{9} - \frac{1}{6}) * (\frac{1}{9}) = (\frac{13}{18}) * (\frac{1}{9})\)
Case 3- We get sum 5 on the third turn.
P(x)=\( (1-\frac{1}{9} - \frac{1}{6})* (1-\frac{1}{9 }- \frac{1}{6})* (\frac{1}{9}) = (\frac{13}{18})^2 * (\frac{1}{9})\)
And so on
Total probability = \(\frac{1}{9}+ (\frac{13}{18}) * (\frac{1}{9}) + (\frac{13}{18})^2 * (\frac{1}{9})+.......infinite terms= (\frac{1}{9})/ (1-\frac{13}{18}) = \frac{2}{5}\)
Probability of getting 7 = \(\frac{6}{36}\) = \(\frac{1}{6}\)
We don't want want sum 7 occurs before sum 5.
Case 1. We get sum 5 on the first turn.
P(x) =\( \frac{1}{9}\)
Case 2- We get sum 5 on the second turn.
\(P(x)= (1-\frac{1}{9} - \frac{1}{6}) * (\frac{1}{9}) = (\frac{13}{18}) * (\frac{1}{9})\)
Case 3- We get sum 5 on the third turn.
P(x)=\( (1-\frac{1}{9} - \frac{1}{6})* (1-\frac{1}{9 }- \frac{1}{6})* (\frac{1}{9}) = (\frac{13}{18})^2 * (\frac{1}{9})\)
And so on
Total probability = \(\frac{1}{9}+ (\frac{13}{18}) * (\frac{1}{9}) + (\frac{13}{18})^2 * (\frac{1}{9})+.......infinite terms= (\frac{1}{9})/ (1-\frac{13}{18}) = \frac{2}{5}\)
Bunuel
