Question

Raj tossed 3 dices and there results are noted down then what is the probability that raj gets 10?

Answer 1

The options for this question are missing:

Following are the options:

a) 1/72

b) 1/9

c) 25/216

d)1/8

In this according to situation there are so many pairs possible

(1,3,6)(1,4,5)(1,5,4)(1,6,3)(2,2,6)(2,3,5)(2,4,4)(2,5,3)(2,6,2)(3,1,6)(3,2,5)(3,3,4)(3,4,3)(3,5,2)(3,6,1)(4,1,5)(4,2,4)(4,3,3)(4,4,2)(4,5,1)(5,1,4)(5,2,3)(5,3,2)(5,4,1)(6,1,3)(6,2,2)(6,3,1)  

6*6*6=216

Total pairs =216

Condition pairs=27

So now 27/216=1/8 will be the answer.

Answer 2

Answer:

$\frac{1}{8}$

Step-by-step explanation:

Raj tossed 3 dices

Total outcomes = $6^n$

Where n is the number of times dice rolled

So, Total outcomes = $6^3 = 216$

Now we are supposed to find the probability that raj gets 10

Favorable outcomes: (1,3,6)(1,4,5)(1,5,4)(1,6,3)(2,2,6)(2,3,5)(2,4,4)(2,5,3)(2,6,2)(3,1,6)(3,2,5)(3,3,4)(3,4,3)(3,5,2)(3,6,1)(4,1,5)(4,2,4)(4,3,3)(4,4,2)(4,5,1)(5,1,4)(5,2,3)(5,3,2)(5,4,1)(6,1,3)(6,2,2)(6,3,1)  = 27

Probability that raj gets 10$=\frac{\text{favorable outcomes}}{\text{total outcomes}}$

                                           $=\frac{27}{216}$

                                           $=\frac{1}{8}$

Hence  the probability that raj gets 10 is  $\frac{1}{8}$