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A pair of dice is tossed twice. What is the probability that
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08 Dec 2017, 20:32
08 Dec 2017, 20:32
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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. \(\frac{7}{18}\)
Drill 2
Question: 8
Page: 525-526
Question: 8
Page: 525-526
Re: A pair of dice is tossed twice. What is the probability that
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29 Dec 2017, 14:03
29 Dec 2017, 14:03
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Explanation
A There are a total of 62 = 36 possibilities for each toss. There are a total of 8 ways we can get a total of 7 or 11 on the first toss: 6 ways to get a total of 7—(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), or (6, 1)—plus 2 ways to get a total of 11—(5, 6) or (6, 5).
Therefore, the probability of getting a total of either 7 or 11 on the first toss is \(\frac{8}{36}=\frac{2}{9}\).
The probability of getting a total of 7 on the second toss is \(\frac{6}{36}=\frac{1}{6}\) so the probability that both of these independent events occur is the product \(\frac{2}{9}\times \frac{1}{6}=\frac{1}{27}\), choice A.
A There are a total of 62 = 36 possibilities for each toss. There are a total of 8 ways we can get a total of 7 or 11 on the first toss: 6 ways to get a total of 7—(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), or (6, 1)—plus 2 ways to get a total of 11—(5, 6) or (6, 5).
Therefore, the probability of getting a total of either 7 or 11 on the first toss is \(\frac{8}{36}=\frac{2}{9}\).
The probability of getting a total of 7 on the second toss is \(\frac{6}{36}=\frac{1}{6}\) so the probability that both of these independent events occur is the product \(\frac{2}{9}\times \frac{1}{6}=\frac{1}{27}\), choice A.
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Re: A pair of dice is tossed twice. What is the probability that
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16 Oct 2022, 09:36
16 Oct 2022, 09:36
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Given that A pair of dice is tossed twice and We need to find 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?
As we are tossing the dice each time => Number of cases in each toss = \(6^2\) = 36
First toss gives a total of either 7 or 11
To get a sum of 7 or 11 following are the possibilities
(1,6), (2,5), (3,4), (4,3), (5,2), (5,6), (6,1), (6,5) => 8 cases
=> P(First Toss giving a total of either 7 or 11) = \(\frac{8}{36}\) = \(\frac{2}{9}\)
The second toss gives a total of 7
To get a sum of 7 following are the possibilities
(1,6), (2,5), (3,4), (4,3), (5,2), (6,1) => 6 cases
=> P(Second Toss giving a total of 7) = \(\frac{6}{36}\) = \(\frac{1}{6}\)
=> Probability that the first toss gives a total of either 7 or 11, and the second toss gives a total of 7 = P(First Toss giving a total of either 7 or 11) * P(Second Toss giving a total of 7)
= \(\frac{2}{9}\) * \(\frac{1}{6}\) = \(\frac{1}{27}\)
So, Answer will be A
Hope it helps!
Watch the following video to learn How to Solve Dice Rolling Probability Problems
As we are tossing the dice each time => Number of cases in each toss = \(6^2\) = 36
First toss gives a total of either 7 or 11
To get a sum of 7 or 11 following are the possibilities
(1,6), (2,5), (3,4), (4,3), (5,2), (5,6), (6,1), (6,5) => 8 cases
=> P(First Toss giving a total of either 7 or 11) = \(\frac{8}{36}\) = \(\frac{2}{9}\)
The second toss gives a total of 7
To get a sum of 7 following are the possibilities
(1,6), (2,5), (3,4), (4,3), (5,2), (6,1) => 6 cases
=> P(Second Toss giving a total of 7) = \(\frac{6}{36}\) = \(\frac{1}{6}\)
=> Probability that the first toss gives a total of either 7 or 11, and the second toss gives a total of 7 = P(First Toss giving a total of either 7 or 11) * P(Second Toss giving a total of 7)
= \(\frac{2}{9}\) * \(\frac{1}{6}\) = \(\frac{1}{27}\)
So, Answer will be A
Hope it helps!
Watch the following video to learn How to Solve Dice Rolling Probability Problems
