Question

two dice are rolled Simultaneously. find the probability of getting (a) sum more than 10 (b) getting a sum more than 6 and a multiple of 3.

Answer 1

total outcome =36

(a)-2/36=1/18

(b)-21/36=7/12.

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Answer 2

Answer:

a) $P = \frac{1}{12}$

b) $P = \frac{1}{12}$

Step-by-step explanation:

Here, total outcomes when two dice are rolled is 36.

Namely, (1,1), (1,2) , (1,3) .........  (5,6) and (6,6)

a) Probability of getting a Sum more than 10 = $\frac{\textrm{Favorable Outcomes}}{\textrm{Total Outcomes}}$

Here, favorable outcomes are (5,6) , (6,5) and (6,6)

So, P(getting a sum more than 10) = $\frac{3}{36}$ = $\frac{1}{12}$

b) Probability of getting a sum more than 6 and a multiple of 3.

So, sum should be either 9 or 12.

Here, favorable outcomes are = (5,4) , (4,5) and (6,6)

Hence, Probability of this event is $\frac{\textrm{Favorable Outcomes}}{\textrm{Total Outcomes}}$

= $\frac{3}{36}  = \frac{1}{12}$