Question

if two ludo dice are thrown once,then what is probability of getting 9 as the sum of the two numbers shown by both the dice

Answer 1

When two dice are thrown simultaneously, all possible outcomes 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.

Let E be the event of getting two numbers whose sum is 9. Then , the favourable outcomes are .

(3,6),(4,5),(5,4),(6,3)

Number of favourable outcomes = 4

Therefore,

P(E) getting two numbers whose sum is 9 = Favourable outcomes/Possible outcomes

= 4/36 = 1/9

Answer 2

Question :- If two balanced dice are tossed once, what is the probability of the event that the sum of the integers coming on the upper sides of the two dice is 9 ?

Solution :-

we know that, when two dice are tossed once, then possible events will be

(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)

Total 36 .

so,

→ Total possible events where sum of upper side of the dice is 9 is = (3,6), (4,5), (5,4) and (6,3) = Total 4 .

therefore,

→ Required Probability = Favourable outcomes / Total number of outcomes = 4/36 = (1/9) (Ans.)

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