Two dices are rolled simultaneously, and their sums are recorded.a) What kind of sample space would exist? b) Determine the theoretical possibility of each of the following events: i) Sum of the...

Two dices are rolled simultaneously, and their sums are recorded.

a) What kind of sample space would exist?

b) Determine the theoretical possibility of each of the following events:

i) Sum of the two dices being divisible by 5.

ii) Rolling a three on at least one die or the sum of the two dies being divisible by four.

iii) A sum that is greater than five.

iv) A sum that is a prime number.

And finally,

c) If you rolled the dice 180 times, how many of each sum would you possibly expect to occur?

Asked on by juno60

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thilina-g's profile pic

thilina-g | College Teacher | (Level 1) Educator

Posted on

a) The sample space would be from 2 to 12. That is sum of the lowest value to sum of the highest value.

The following  is sample space.

Value            2  3  4  5  6  7  8  9  10  11  12

Frequency     1  2  3  4  5  6  5  4   3   2    1

If you analyse the possible combinations of the two dices, you can find the frequency as above.

The total number of possibilities = sum of frequency

`T = 1+2+3+4+5+6+5+4+3+2+1`

`T = 36`

 There are 36 possibillities altogether.

b)

(i) Sum of the two dices being divisible by 5.

The values that can be divided by 5 are 5 and 10. The possibilites this can happen is 4 + 3 , which is 7.

Therefore probability that sum is divisible by 5 is `= 7/36`

(ii) Probability of not rolling three on any of the dice `= 5/6 xx 5/6`

                                                                          `= 25/36`

Therefore probability of rolling three on at least one dice

                                                                          `= 1 - 25/36`

                                                                          `= 11/36`


The probability of sum of two dices divisible by 4 ` = (3+5+1)/36`

                                                                      `= 9/36`

                                                                      `= 1/4`

Therefore the probability of rolling three on at least one dice or sum of two dices divisible by 4 is,

                                          `= 11/36+9/36`

                                          `= 20/36`

                                         ` = 5/9`

 (iii) A sum that is greater than five.

The number of possibilities this can happen is

                      `= 5 + 6 + 5 + 4 + 3 + 2 + 1`

                      `= 26`

Therefore the probability that sum is greater than 5 `= 26/36`

                                                                          `= 13/18`

(iv) A sum that is a prime number.

The primer number in the sample space are 2, 3, 5, 7, 11

Therefore the number of possibilities this can happen =

                      `= 1+2+4+6++2`

                      `= 15`

 Therefore the probability that sum is a prime number `= 15/36`

                                                                            `= 5/12`

(v) The frequency in the above table for is for rolling the two dices one time.

To find the probability of each sum you can divide by 36.

Therefore probability of 2, P(2)= 1/36

In similar way you can find the other probabilities also.

To find the occurences of each sum if dices are rolled 180 times, you can simply multiply the by 180.

Number of occurences of 2 = 1/36 x 180 = 5.

Like wise you can get the other answers too.

thilina-g's profile pic

thilina-g | College Teacher | (Level 1) Educator

Posted on

I will continue the part c here. I have put it wrong their as (v).

C) The frequency in the above table for is for rolling the two dices one time.

To find the probability of each sum you can divide by 36.

Therefore probability of 2, P(2)= 1/36

Likewise,

P(3) = 2/36

P(4) = 3/36

P(5) = 4/36

P(6) = 5/36

P(7) = 6/36

P(8) = 5/36

P(9) = 4/36

P(10) = 3/36

P(11) = 2/36

P(12) = 1/36

To find the occurences of each sum if dices are rolled 180 times, you can simply multiply the by 180.

Number of occurences of 2 N(2) = 1/36 x 180 = 5.

Like wise you can get the other answers too.

N(3) = 10

N(4) = 15

N(5) = 20

N(6) = 25

N(7) = 30

N(8) = 25

N(9) = 20

N(10) = 15

N(11) = 10

N(12) = 5

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