Question

Two dice are thrown simultaneously. Find the probability of getting a multiple of 2 on one die and a multiple of 3 on the other die.

Answer 1

Two dice are thrown simultaneously

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 : getting multiple of 2 on one die and multiple of 3 on other die
A={(2,3)(2,6)(4,3)(4,6) (6,3)(6,6)(6,2)(6,4)(3,2)(3,6)}
n(A)=10
p(A)=n(A)/n(s)
=10/36
=5/18

Answer 2

Answer:

Total outcomes=36

Let E= Event of getting a multiple of 2 on one dice and a multiple of 3 in another dice.

Here the multiple of 2 are 2,4,6 and multiple of 3 are 3,6 .

So favourabl outcomes for E are (2,3) (4,3) (6,3) (2,6) (4,6) (6,6) (3,2) (3,4) (3,6) (6,2),(6,4)

~ number of outcomes favourable to E =11

Therefore

required probability =p(E)

=11/36

so. and is 11/36

Step-by-step explanation: