Two dice are rolled simultaneously. Find the probability ofGetting the...
Here n(S) = 6 x 6 = 36
E={(1,2),(1,5),(2,1),(2,4),(3,3),(3,6),(4,2),(4,5),(5,1),(5,4),(6,3) ,(6,6),(1,3),(2,2),(2,6),(3,1),(3,5), (4,4),(5,3),(6,2)}
=> n(E)=20
Required Probability n(P) = n(E)/n(S) = 20/36 = 5/9.
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Two dice are rolled simultaneously. Find the probability ofGetting the...
When two dice are rolled then u know that what will be the sample events and the total possible outcomes are 6×6= 36
favourable outcomes of the events i.e, getting the sum of no. on two Faces divisible by 3 or 4 = ( 2,1) ( 3,1) ( 5,1) ( (1,2) ( 2,2) (4,2) (6,2) (1,3) (3,3) (5,3) (6,3) (2,4) (4,4) (5,4) (1,5) (3,5) (4,5) (2,6) (3,6) (6,6) = 20
so p( getting the sum of no. on two Faces divisible by 3or 4) = favourable outcomes/ possible outcomes
= 20/36 ,
= 5/9 ans
Two dice are rolled simultaneously. Find the probability ofGetting the...
Here, n (S) = 6 × 6 = 36.
E= { (1,2) , (1,5) , (2,1), (2,4), (3,3), (3,6), (4,2), (4,5), (5,1), (5,4), (6,3), (6,6), (1, 3), (2,2), (2,6), (3,1), (3, 5), (4,4), (5, 3), (6, 2) }
n (E) = 20
Therefore, required probability, n (P) = n (E)/n (S) = 20/36 = 5/ 9.
Hence, answer is justified...
Hope it helps!!!!
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