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In a simultaneous throw of a pair of dice, find the probability of getting a total of 9 or more.
When a pair of dice are thrown simultaneously the maximum sum possible is 12.
The possible cases of getting a sum of 9 are:
Β Β Β Β Β Β Β Β Β Β dice 1Β Β Β dice 2
Β Β Β Β Β Β Β Β Β Β fourΒ Β Β Β Β five.....................(1)
Β Β Β Β Β Β Β Β Β Β fiveΒ Β Β Β Β four.....................(2)
Β Β Β Β Β Β Β Β Β Β Β sixΒ Β Β Β Β three....................(3)
Β Β Β Β Β Β Β Β Β Β threeΒ Β Β Β Β six.......................(4)Β Β
So a total of 4 cases are possible
The possible cases of getting a sum of 10 are:
Β Β Β Β Β Β Β Β Β Β dice 1Β Β Β dice 2
Β Β Β Β Β Β Β Β Β Β Β fourΒ Β Β Β six......................(1)
Β Β Β Β Β Β Β Β Β Β Β sixΒ Β Β Β Β four.....................(2)
Β Β Β Β Β Β Β Β Β Β Β fiveΒ Β Β Β five.....................(3)
So a total of 3 cases are possible
The possible cases of getting a sum of 11 are:
Β Β Β Β Β Β Β Β Β Β dice 1Β Β Β Β dice 2
Β Β Β Β Β Β Β Β Β Β fiveΒ Β Β Β Β Β six...................(1)
Β Β Β Β Β Β Β Β Β Β Β sixΒ Β Β Β Β Β five..................(2)
So a total of 2 cases are possible
The possible cases of getting a sum of 12 are:
Β Β Β Β Β Β Β Β Β dice 1Β Β Β dice 2
Β Β Β Β Β Β Β Β Β Β sixΒ Β Β Β Β Β six.....................(1)
So a total of 1 case is possible.
Total number ways of getting a sum of 9 or more = 4 + 3 + 2 + 1 = 10 ways
Total outcomes when two dice are rolled simultaneously = 6 * 6 = 36 ways
Probability of getting a sum of 9 or more = Expected number of outcomes/ Total outcomes
= $$\frac{10}{36}$$
= $$\frac{5}{18}$$
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