Question 18

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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