Two different dice are thrown together. Find the probability that the numbers obtained have (1)even sum and (2)even product
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Solution:-
If two different dice are thrown together, they have numbers 1, 2, 3, 4, 5, 6 and 1, 2, 3, 4, 5, 6 on them.
Total number of outcomes -
(1,1) ; (1,2); (1,3) ; (1,4) ; (1,5) ; (1,6) ; (2,1) ; (2,2) ; (2,3) ; (2,4) ; (2,5) ; (2,6) ; (3,1) ; (3,2) ; (3,3) ; (3,4) ; (3,5) ; (3,6) ; (4,1) ; (4,2) ; (4,3) ; (4,4) ; (4,5) ; (4,6) ; (5,1) ; (5,2) ; (5,3) ; (5,4) ; (5,5) ; (5,6) ; (6,1) ; (6,2) ; (6,3) ; (6,4) ; (6,5) ; (6,6) = 36
Question 1 : Sum of two even numbers and sum of two odd numbers will be an even number, So
Probability of getting even sum :
P(ES) = (1,1) ; (1,3) ; (1,5) ; (2,2) ; (2,4) ; (2,6) ; (3,1) ; (3,3) ; (3,5) ; (4,2) ; (4,4) ; (4,6) ; (5,1) ; (5,3) ; (5,5) ; (6,2) ; (6,4) ; (6,6)
= 18/36
= 1/2
Question 2 : Product of two even numbers and product of an even number and an odd number will be an even number, So
Probability of getting even product :
P(EP) = (1,2) ; (1,4) ; (1,6) ; (2,1) ; (2,2) ; (2,3) ; (2,4) ; (2,5) ; (2,6) ; (3,2) ; (3,4) ; (3,6) ; (4,1) ; (4,2) ; (4,3) ; (4,4) ; (4,5) ; (4,6) ; (5,2) ; (5,4) ; (5,6) ; (6,1) ; (6,2) ; (6,3) ; (6,4) ; (6,5) ; (6,6)
= 27/36
= 3/4
Answer.
If two different dice are thrown together, they have numbers 1, 2, 3, 4, 5, 6 and 1, 2, 3, 4, 5, 6 on them.
Total number of outcomes -
(1,1) ; (1,2); (1,3) ; (1,4) ; (1,5) ; (1,6) ; (2,1) ; (2,2) ; (2,3) ; (2,4) ; (2,5) ; (2,6) ; (3,1) ; (3,2) ; (3,3) ; (3,4) ; (3,5) ; (3,6) ; (4,1) ; (4,2) ; (4,3) ; (4,4) ; (4,5) ; (4,6) ; (5,1) ; (5,2) ; (5,3) ; (5,4) ; (5,5) ; (5,6) ; (6,1) ; (6,2) ; (6,3) ; (6,4) ; (6,5) ; (6,6) = 36
Question 1 : Sum of two even numbers and sum of two odd numbers will be an even number, So
Probability of getting even sum :
P(ES) = (1,1) ; (1,3) ; (1,5) ; (2,2) ; (2,4) ; (2,6) ; (3,1) ; (3,3) ; (3,5) ; (4,2) ; (4,4) ; (4,6) ; (5,1) ; (5,3) ; (5,5) ; (6,2) ; (6,4) ; (6,6)
= 18/36
= 1/2
Question 2 : Product of two even numbers and product of an even number and an odd number will be an even number, So
Probability of getting even product :
P(EP) = (1,2) ; (1,4) ; (1,6) ; (2,1) ; (2,2) ; (2,3) ; (2,4) ; (2,5) ; (2,6) ; (3,2) ; (3,4) ; (3,6) ; (4,1) ; (4,2) ; (4,3) ; (4,4) ; (4,5) ; (4,6) ; (5,2) ; (5,4) ; (5,6) ; (6,1) ; (6,2) ; (6,3) ; (6,4) ; (6,5) ; (6,6)
= 27/36
= 3/4
Answer.
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