Question 1201645: Find the probability that the sum is as stated when a pair of dice is rolled. (Enter your answers as fractions.)
(a) odd and doubles
(b) odd or doubles
Found 2 solutions by Edwin McCravy, mananth:
Answer by Edwin McCravy(20054)   (Show Source):
You can put this solution on YOUR website!
Here are all 36 possible rolls:
(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) odd and doubles
Count how many BOTH have odd sums AND ALSO are doubles. Make a fraction with
that number as the numerator and 36 as the denominator. If it will reduce, do
so. (b) odd or doublesCount how many EITHER have odd sums OR are doubles. Make a fraction with that
number as the numerator and 36 as the denominator. If it will reduce, do so.
Edwin
Answer by mananth(16946)  (Show Source):
You can put this solution on YOUR website! Answer on Question #66364 - Math - Statistics and ProbabilityQuestionFind the probability that the sum is as stated when a pair of dice is rolled. (Enter your answers as fractions.)(a)odd and doubles(b)odd or doubles
a) Sample space = Total number of outcomes = 36
Event of getting sum odds
A= (1,2),(1,4),(1,6),(2,1),(2,3),(2,5),(3,2).(3,4),
(3,6),(4,1),(4,3),(4,5),(5,2),(5,4),(5,6)(6,1),(6,3),(6,5)
= 18
Doubles =(1,1)(2,2)(3,3),(4,4),(5,5),(6,6)
=(6)
P (odd and double) = (0)
b) P(odd or doub;es ) = (18+6)/36 = 24/36= 2/3
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