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You have two normal, fair, dice, with faces labelled $1,2, \dots ,6.$ If you throw both dice, which of the following is true about the total value shown by the dice?

- The probability that the total is $6$ is less than the probability that the total is $9.$
- The probability that the total is $6$ is equal to the probability that the total is $9.$
- The probability that the total is $6$ is greater than the probability that the total is $9.$
- None of the above.

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Best answer

Possible combination to get $6$ as sum $= (2,4) (4,2) (3,3) (5,1) (1,5)$

Possible combination to get $9$ as sum $= (4,5) (5,4) (6,3) (3,6)$

The probability that the total is $6 =\frac{5}{36}=0.138$

The probability that the total is $9 =\frac{4}{36}=0.111$

Here,

The probability that the total is $6 >$ The probability that the total is $9$

$0.138 > 0.111$

Hence,Option(C)The probability that the total is 6 is greater than the probability that the total is $9.$

Possible combination to get $9$ as sum $= (4,5) (5,4) (6,3) (3,6)$

The probability that the total is $6 =\frac{5}{36}=0.138$

The probability that the total is $9 =\frac{4}{36}=0.111$

Here,

The probability that the total is $6 >$ The probability that the total is $9$

$0.138 > 0.111$

Hence,Option(C)The probability that the total is 6 is greater than the probability that the total is $9.$