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.
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