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When two dice are rolled, 36 equally likely outcomes are possible as shown below. (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6) (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) Let x be the sum of the two numbers. Let P be the probability of the desired outcome. Compare the following charts and determine which chart shows the probability distribution for the sum of the two numbers.

When two dice are rolled, 36 equally likely outcomes are possible as shown below. (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6) (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) Let-example-1
User Bitxwise
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2 Answers

4 votes

Answer: B

Explanation:

User Stevemac
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Given that When two dice are rolled, 36 equally likely outcomes are possible as shown below. (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6) (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)

The sum of the 36 outcomes are as follows: 2, 3, 4, 5, 6, 7, 3, 4, 5, 6, 7, 8, 4, 5, 6, 7, 8, 9, 5, 6, 7, 8, 9, 10, 6, 7, 8, 9, 10, 11, 7, 8, 9, 10, 11, 12

The number of the sums are as follows:
n(2) = 1
n(3) = 2
n(4) = 3
n(5) = 4
n(6) = 5
n(7) = 6
n(8) = 5
n(9) = 4
n(10) = 3
n(11) = 2
n(12) = 1

The probability of the sums are given as follows:
p(2) = 1/36
p(3) = 2/36 = 1/18
p(4) = 3/36 = 1/12
p(5) = 4/36 = 1/9
p(6) = 5/36
p(7) = 6/36 = 1/6
p(8) = 5/36
p(9) = 4/36 = 1/9
p(10) = 3/36 = 1/12
p(11) = 2/36 = 1/18
p(12) = 1/36

Therefore, the chart that shows the probability distribution for the sum of the two numbers is chart B.
User Rahul Soni
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