Given : two dice are tossed once
To find : Probability of getting an even number on the first die or a total 8.
The formula used : Probability
= \(\frac{favourable\,number\,of\,outcomes}{Total\,no.of\,outcomes}\)
P(A or B) = P(A) + P(B) - P(A and B)
A die is numbered from 1 to 6
When two dice are tossed once, total number of outcomes = 62 = 36
Let A denote the event of getting an even number on the first die and B denote the event of getting a total of 8
For getting an even number on the first die
Favourable outcomes =
{(2,1) ,(2,2) ,(2,3) ,(2,4) ,(2,5) ,(2,6) ,(4,1) ,(4,2) ,(4,3) ,(4,4) ,(4,5) ,(4,6) , (6,1) ,(6,2) ,(6,3) ,(6,4) ,(6,5) ,(6,6) }
Favourable number of outcomes = 18
Probability of getting an even number on the first die = P(A) = \(\frac{18}{36}\)
For getting a total of 8
Favourable outcomes =
{ (2,6) , (4,4) , (6,2) ,(5,3) , (3,5) }
Favourable number of outcomes = 5
Probability of getting a total of 8 = P(A) = \(\frac{5}{36}\)
For getting an even number on the first die and a total of 8
Favourable outcomes = {(2,6) , (4,4) , (6,2)}
Probability of getting an even number on the first die and a total of 8 =
P(A and B) = \(\frac{5}{36}\)
P(A or B) = P(A) + P(B) - P(A and B)
P(A or B) = \(\frac{18}{36}+\frac{5}{36}-\frac{3}{36}\)
P(A or B) = \(\frac{18+5-3}{36}\) = \(\frac{20}{36}\) = \(\frac{5}{9}\)
Probability of getting an even number on the first die or a total 8 = \(\frac{5}{9}\)
