Two six-sided dice are tossed. The sample space is given...

Two six-sided dice are tossed. The sample space is given below. Round all answers to 3 decimal places. 1,1 2,1 3,1 4,1 5,1 6,1 1,2 2,2 3,2 4,2 5,2 6,2 1,3 2,3 3,3 4,3 5,3 6,3 1,4 2,4 3,4 4,4 5,4 6,4 1,5 2,5 3,5 4,5 5,5 6,5 1,6 2,6 3,6 4,6 5,6 6,6 (a) What is the probability of rolling a 3 on either dice? (b) What is the probability of rolling a 5 and 7? (c) What is the probability of rolling a 2 and an odd number (in either order)? (d) What is the probability of rolling two numbers whose sum is odd or a multiple of 3?

Transcript

00:01 It is given that two six -sided dice are rolled and their outcomes are given below.

00:10 Thus, the total number of sample space for the two six -sided rolls, dice rolled is 36.

00:20 Now, in part a, we have to find the probability of having three on either of the two die.

00:30 This implies we have to find the probability of having 3 in the first die union, the probability of having 3 also in the second die.

00:43 Now, the number of outcomes where 3 appear in the first die is for this 6 outcomes 3 appear in the first day.

00:57 Thus there are six outcomes where three appear in the first day and the number of outcomes where three appear in the second die is for this six outcomes the number three appear in the second die.

01:16 Thus the number of outcomes where three appear in the second die is six and the number of outcomes where three appear in the second die is six and the number of outcomes where three appear in the first day.

01:27 Die and also three appear in the second die is this one outcome thus the number of outcomes where three appear in both the die is one now the required probability is calculated as probability of having three in the first die union the probability of having three in the second die which is having the three and either die is calculated as probability of having three 3 in the 1st die plus probability of having 3 in the 2nd die minus probability of having 3 in the both of the dies.

02:08 And this is calculated as this probability is obtained by the number of outcomes where 3 appear in the 1st die divided by the total number of sample space of 36.

02:20 Plus for the second day the number of favorable outcomes is 6 divided by the sample space of 36 minus for this the number of favorable outcome is 1 divided by 36 now simplifying this gives the 0 .306 thus this is the required probability now in part b we have to find the probability of having 5 in the first day and 7 in the second day as we know that there are only six -sided in the dice, the probability of having the number 7 is not possible, and which is represented by the probability of 0.

03:04 Now in part c, we have to find the probability of having 2 and an odd number in the rolling...

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