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Two dice, one blue and one grey, are thrown at the same time. Now
(i) Complete the following table:
(ii) A student argues that-there are 11 possible outcomes 2
Given:
Two dice, one blue and one grey, are thrown at the same time.
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 $\frac{1}{11}$.
To do:
We have to
(i) complete the given table.
(ii) We have to find whether the given argument is true.
Solution:
When two dice (one blue and one green) are thrown at the same time, the total number of outcomes $=6 \times 6=36$
This implies,
The total number of possible outcomes $n=36$.
(i) When the sum on the two dice $=2$, the possible outcome is $(1,1)$
Number of favorable outcomes $=1$
Probability that the sum on the two dice is 2 $=\frac{1}{36}$
When the sum on the two dice $=3$, the possible outcome is $(1,2), (2,1)$
Number of favorable outcomes $=2$
Probability that the sum on the two dice is 3 $=\frac{2}{36}$
When the sum on the two dice $=4$, the possible outcome is $(1,3), (2,2), (3,1)$
Number of favorable outcomes $=3$
Probability that the sum on the two dice is 4 $=\frac{3}{36}$
When the sum on the two dice $=5$, the possible outcome is $(1,4),(2,3),(3,2),(4,1)$
Number of favorable outcomes $=4$
Probability that the sum on the two dice is 5 $=\frac{4}{36}$
When the sum on the two dice $=6$, the possible outcome is $(1,5),(2,4),(3,3),(4,2),(5,1)$
Number of favorable outcomes $=5$
Probability that the sum on the two dice is 6 $=\frac{5}{36}$
When the sum on the two dice $=7$, the possible outcome is $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$
Number of favorable outcomes $=6$
Probability that the sum on the two dice is 7 $=\frac{6}{36}$
When the sum on the two dice $=8$, the possible outcome is $(2,6),(3,5),(4,4),(5,3),(6,2)$
Number of favorable outcomes $=5$
Probability that the sum on the two dice is 8 $=\frac{5}{36}$
When the sum on the two dice $=9$, the possible outcome is $(3,6),(4,5),(5,4),(6,3)$
Number of favorable outcomes $=4$
Probability that the sum on the two dice is 9 $=\frac{4}{36}$
When the sum on the two dice $=10$, the possible outcome is $(4,6),(5,5),(6,4)$
Number of favorable outcomes $=3$
Probability that the sum on the two dice is 10 $=\frac{3}{36}$
When the sum on the two dice $=11$, the possible outcome is $(5,6),(6,5)$
Number of favorable outcomes $=2$
Probability that the sum on the two dice is 11 $=\frac{2}{36}$
When the sum on the two dice $=12$, the possible outcome is $(6,6)$
Number of favorable outcomes $=1$
Probability that the sum on the two dice is 12 $=\frac{1}{36}$
No, the outcomes are not equally likely.
(ii) No, the outcomes are not equally likely from the above table we see that, there is different probability for different outcomes.
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