Question

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Answer 1

A dice has six numbers written on six faces. We are throwing two dice, one blue coloured and one grey coloured.


When a dice is thrown, we can get any number from 1 to 6.


We are throwing the two dice simultaneously. We will denote an outcome in the following format:

$(\text{Number on Blue Dice , Number on Grey Dice})$

For example, if we get 3 on Blue Dice, and 5 on Grey Dice, we will denote the outcome as (3,5).

There are 6 possible outcomes on each dice. Since we are throwing 2 dice, the total number of possible outcomes will be $6\times 6 = 36$


All the possible outcomes are as follows:


$\boxed{\begin{array}{ccccccccccc} (1,1) & , & (1,2) & , & (1,3) & , & (1,4) & , & (1,5) & , & (1,6) \\ \\ (2,1) & , & (2,2) & , & (2,3) & , & (2,4) & , & (2,5) & , & (2,6) \\ \\ (3,1) & , & (3,2) & , & (3,3) & , & (3,4) & , & (3,5) & , & (3,6) \\ \\ (4,1) & , & (4,2) & , & (4,3) & , & (4,4) & , & (4,5) & , & (4,6) \\ \\ (5,1) & , & (5,2) & , & (5,3) & , & (5,4) & , & (5,5) & , & (5,6) \\ \\ (6,1) & , & (6,2) & , & (6,3) & , & (6,4) & , & (6,5) & , & (6,6) \end{array}}$


So, we see that:

Total Number of Possible Outcomes = 36



Now, we have to find the probability of certain events. For each event, we need to find the number of outcomes favourable to that event.

1) Sum of Numbers on Dice is 8

Suppose this is Event A.

The Favourable outcomes are:

(2,6) , (3,5) , (4,4) , (5,3) , (6,2)

So, Number of favourable outcomes = 5

So, Probability is:

$\boxed{P(A) = \frac{5}{36}}$


2) Sum of Numbers on Dice is 13

Suppose this is Event B.

The maximum sum possible is 6+6=12. So there are no outcomes with the sum as 13.

So, Number of favourable outcomes = 0


So, Probability is:

$P(B) = \frac{0}{36} \\ \\ \\ \implies \boxed{P(B) = 0}$


3) Sum of Numbers on Dice is less than or equal to  12

Suppose this is Event C.

The maximum sum possible is 6+6=12. 

So, all outcomes have the sum less than or equal to 12.

So, Number of favourable outcomes = 36 (All outcomes)


So, Probability is:

$P(C) = \frac{36}{36} \\ \\ \\ \implies \boxed{P(C) = 1}$