Question

two different dice are tossed together, find the probability (i) of getting a doublet, (ii) of getting a sum of 10, the number on the two dice.

Answer 1

Total no.of possible outcomes = 36
(i) Outcomes favourable for getting a doublet are (1,1), (2,2), (3,3), (4,4), (5,5), (6,6)
Favourable no.of events = 6
Probability of getting doublet = 6/36 =1/6

(ii) Outcomes favourable for getting sum of 10 are (4,6), (5,5), (6,4)
Favourable no. of events = 3
Probability of getting sum of 10 = 3/36 =1/12
(iii) Unable to understand this part of ques.

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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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