Question

two different dice are tossed together find the probability
i) off getting a doubled
ii) off getting a sum of number 10 on the two dice

Answer 1

Solution :-

Possible outcomes when two dice are tossed together are as follows.

⇒ (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)

(1) Probability of getting a doubled - 

Favourable outcomes = (1, 1); (2, 2) ; (3, 3); (4, 4) ; (5, 5) ; (6, 6)

⇒ 6/36

= 1/6

(2) Probability of getting a sum of 10 -

Favourable outcomes = (4, 6); (5, 5) ; (6, 4)

⇒ 3/36

= 1/12

Answer.

Answer 2

A die has 6 faces marked as 1,2, 3, 4, 5, 6. When we throw a die , than total number of outcomes = 6 { 1 , 2 , 3 , 4 , 5 , 6 }
But when we throw two dice simultaneously , than total number of outcomes = 6² = 6 × 6 = 36
Total possible outcomes on throwing two dice are:
{(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)}
Number of all possible outcomes=36

SOLUTION:
i)Total number of outcomes= 36
Favourable outcomes are : (1,1),(2,2) ,(3,3),(4,4),(5,5),(6,6)
Number of outcomes favourable = 6
Probability = Number of favourable outcomes / Total number of outcomes.
Required probability = P(getting the doublet) = 6/36 = ⅙

Hence, the probability of getting the doublet = ⅙.

ii)Total number of outcomes= 36
Favourable outcomes are : (5,5),(4,6) ,(6,4)
Number of outcomes favourable = 3
Probability = Number of favourable outcomes / Total number of outcomes.
Required probability = P(getting the sum on both the die as 10) = 3 /36 = 1/12

Hence, the probability of getting the sum on both the die as 10 = 1/12.

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