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27 May 2018, 12:20
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
71% (02:08) correct
29%
(02:36)
wrong
based on 367
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A pair of dice is tossed twice. What is the probability that the first toss gives a total of either 7 or 11 and the second toss gives a total of 7 ?
A. \(\frac{1}{27}\)
B. \(\frac{1}{18}\)
C. \(\frac{1}{9}\)
D. \(\frac{1}{6}\)
E. \(\frac{7}{18}\)
A. \(\frac{1}{27}\)
B. \(\frac{1}{18}\)
C. \(\frac{1}{9}\)
D. \(\frac{1}{6}\)
E. \(\frac{7}{18}\)
27 May 2018, 13:09
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Bunuel
total number of sample space in each throw = 36
number of outcomes in first toss = (1,6) (2,5) (3,4) (4,3) (5,2) (6,1) (5,6) (6,5)
probability of getting 7 or 11 in first toss = 8 / 36
number of outcomes in second toss = (1,6) (2,5) (3,4) (4,3) (5,2) (6,1) = 6
probability of getting 7 in second toss = 6/36 = 1/36
so the probability of getting 7 or 11 in first toss and 7 in second toss = 8 / 36 * 1/6 = 1 /27
so the answer should be A (not sure)
28 May 2018, 00:09
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Solution
Given:
- • A pair of dice is tossed twice
To find:
- • The probability that the first toss gives a total of either 7 or 11 and the second toss gives a total of 7
Approach and Working:
- • The number of ways the total becomes 7 = (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) = 6 ways
• The number of ways the total becomes 11 = (5, 6), (6, 5) = 2 ways
- • Probability of getting a 7 in the first toss = \(\frac{6}{36} = \frac{1}{6}\)
• Probability of getting a 11 in the first toss = \(\frac{2}{36} = \frac{1}{18}\)
• Probability of getting a 7 or 11 in the first toss = \(\frac{1}{6} + \frac{1}{18} = \frac{2}{9}\)
• Probability of getting a 7 in the second toss = \(\frac{6}{36} = \frac{1}{6}\)
Hence, the correct answer is option A.
Answer: A
