Question

A die is rolled three times. The probability of getting their sum equal to a prime number of the form 4n+1 is

Answer 1

Answer:

5/36

Explanation:

Answer 2

Therefore The probability of getting their sum equal to a prime number of form 4n+1 is 7/72.

Given:

A die is rolled 3 times.

To Find:

The probability of getting their sum equal to a prime number of the form 4n+1.

Solution:

The given question can be solved very easily as shown below.

Given that a die is rolled 3 times,

Total number of possible outcomes = 6³

If n=0 → 4(0) + 1 = 1

⇒  Getting sum '1' is impossible with 3 dice.

If n=1 → 4(1) + 1 = 5 → '5' is a prime number.

⇒ Possible outcomes = ( 1,1,3 ); ( 1,3,1 ); ( 3,1,1 ); (1,2,2); (2,1,2); (2,2,1)

⇒ Number of possible outcomes to get sum '5' = 6 ways

If n=2 → 4(2) + 1 = 9 → '9' is not a prime number.

If n=3 → 4(3) + 1 = 13 → '13' is a prime number.

⇒ Possible outcomes = (6,6,1); (6,1,6); (1,6,6); (5,5,3); (5,3,5); ( 3,5,5); (6,4,3); (6,3,4); (3,6,4); (3,4,6); (4,3,6); (4,6,3);

⇒ Number of possible outcomes to get sum '13' = 12 ways

If n=4 → 4(4) + 1 = 17 → '17' is a prime number.

⇒ Possible outcomes = (6,6,5); (6,5,6); (5,6,6)

⇒ Number of possible outcomes to get sum '17' = 3 ways

We can't get a sum of more that 18 with 3 dice so the above-stated are the total possible outcomes.

⇒ Total number of favorable outcomes = 6 + 12 + 3 = 21 ways

⇒ The probability of getting their sum equal to a prime number of the form 4n+1 = ( Total number of favorable outcomes ) / ( Total number of possible outcomes )

⇒ The probability of getting their sum equal to a prime number of form 4n+1 = 21/216 = 7/72

Therefore The probability of getting their sum equal to a prime number of form 4n+1 is 7/72.

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