Question

two dice are tossed find the probability (a) sum of the digits on upper face is prime number. (b) digits on the upper face of the first die is less than the digit on the second die​

Answer 1

Answer:

p(A)=5/12 and p(B)=5/12

Step-by-step explanation:

S={(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)}

n(S)=36

Event A: Sum of the digits on upper face is prime number.

A={ (1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4,3),(5, 2), (5, 6), (6, 1), (6, 5) }

n(A) = 15.

p(A)=n(A)/n(S)

=15/36

=5/12

Event B: Digits on the upper face of the first die is less than the digit on the second die​.

B={ (1,2),(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6),(4,5),(4,6),(5,6)}

n(B)=15.

p(B)=n(B)/n(S)

=15/36

=5/12

I hope you find it useful.

Answer 2

Answer:

correct answer is 5/12

5/12