**Probability ncert solutions Chapter 15 Exercise 15.1 Question 22**

**22.** Two dice, one blue and one grey are thrown at the same time. What is the probability of different sums that can appear on the top of different dice.

Also, a student argues that "there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability . Do you agree with this argument? Justify your answer.

**Solution:**

Sum equal to 2 can appear in just 1 way which is (1, 1) (We have 1 appearing on both of the dice).

Sum equal to 3 can appear in 2 ways which are (1, 2) and (2, 1)

Sum equal to 4 can appear in 3 ways which are (1, 3), (3, 1) and (2, 2)

Sum equal to 5 can appear in 4 ways which are (2, 3), (3, 2), (1, 4) and (4, 1)

Sum equal to 6 can appear in 5 ways which are (1, 5), (5, 1), (2, 4), (4, 2) and (3, 3)

Sum equal to 7 can appear in 6 ways which are (1, 6), (6, 1), (2, 5), (5, 2), (3, 4) and (4, 3)

Sum equal to 8 can appear in 5 ways which are (2, 6), (6, 2), (3, 5), (5, 3) and (4, 4)

Sum equal to 9 can appear in 4 ways which are (3, 6), (6, 3), (4, 5) and (5, 4)

Sum equal to 10 can appear in 3 ways which are (4, 6), (6, 4) and (5, 5)

Sum equal to 11 can appear in 2 ways which are (5, 6) and (6, 5)

Sum equal to 12 can appear in 1 way which is (6, 6)

Therefore, we can find probabilities of all the sums which can appear on the top of two dice.

Total number of possible outcomes =1+2+3+4+5+6+5+4+3+2+1=36

A student argues that "there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability .

We do not agree with this argument because there are different number of possible outcomes for each sum. According to the above table, we can see that each sum has different probability.