Math, asked by shree4424, 1 year ago

First 6 faced die which is numbered 1 through 6is thrown then a 5 faced die which is numbered1 through 5 is thrown. What is the probabilitythat sum of the numbers on the upper facesof the dice is divisible by 2 or 3?​

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Answered by 18shreya2004mehta
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Answered by ChitranjanMahajan
1

Probability that the sum of the numbers on the upper faces of the dice be divisible by 2 or 3 is equal to 19 / 30.

• The maximum number on a six-faced die is 6, and that on a five-faced die is 5.

• Total number of possible outcomes on throwing both the dice = 6 × 5 = 30

• The maximum sum of numbers on the upper faces of the two dice could be 6 + 5 = 11.

• The numbers divisible by 2 within 11 are 2, 4, 6, 8, and 10. Therefore, the sum of the numbers on the upper faces of both the dice should be 2, 4, 6, 8, or 10.

• The possible arrangements with maximum 6 on the first die and 5 on the second die are :

For the sum to be 2 - (1,1)

For the sum to be 4 - (1,3) , (2,2) , (3,1)

For the sum to be 6 - (1,5) , (2,4) , (3,3) , (4,2) , (5,1)

For the sum to be 8 - (3,5) , (4,4) , (5,3) , (6,2)

For the sum to be 10 - (5,5)

• Therefore, total number of favourable outcomes for the sum to be divisible by 2 = 14

• The numbers divisible by 3 within 11 are 3, 6, 9. Therefore, the sum of the numbers on the upper faces of both the dice should be 3, 6, or 9.

• The possible arrangements with maximum 6 on the first die and 5 on the second die are :

For the sum to be 3 - (1,2) , (2,1)

For the sum to be 6 - Same combinations as that stated above for the sum to be 2. Hence, need not be considered twice.

For the sum to be 9 - (4,5) , (5,4) , (6,3)

• Therefore, total number of favourable outcomes for the sum to be divisible by 3 (excluding 6) = 5

• Probability of getting a sum on throwing the dice that is either divisible by 2 or 3 (P) = (Number of favourable outcomes for getting 2 as the sum + Number of favourable outcomes for getting 3 as the sum) / Total number of possible outcomes

=> P = (14 + 5) / 30

=> P = 19 / 30

Probability of getting the sum as 2 or 3 = 19 / 30

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