# 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

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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.

Updated on 10-Oct-2022 13:26:59