Question

Two dice, one blue and one grey are thrown at the same time. Complete the following table:

Event: Sum on two dice'	2	3	4	5	6	7	8	9	10	11	12
Probability

From the above table 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 of $\frac{1}{11}$. Do you agree with this argument?

Answer 1

In a throw of pair of dice, blue and grey, total no of possible outcomes$=36(6\times 6)$ which are

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

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

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

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

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

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

$E⟶$ event of getting sum on $2$ dice as $2$

No. of favorable outcomes $=1(i.e.,(1,1\left)\right)$

Probability, $P\left(E\right)=\frac{No.ofFavorableOutcomes}{TotalNo.ofPossibleOutcomes}$

$P\left(E\right)=\frac{1}{36}$

$E⟶$ event of getting sum as $3$

No. of favorable outcomes $=2(i.e.,(1,2\left)\right(2,1\left)\right)$

$P\left(E\right)=\frac{2}{36}=\frac{1}{18}$

$E⟶$ event of getting sum as $4$

No. of favorable outcomes $=3(i.e.,(3,1\left)\right(2,2\left)\right(1,3\left)\right)$

$P\left(E\right)=\frac{3}{36}=\frac{1}{12}$

$E⟶$ event of getting sum as $5$

No. of favorable outcomes $=4(i.e.,(1,4\left)\right(2,3\left)\right(3,2\left)\right(4,1\left)\right)$

$P\left(E\right)=\frac{4}{36}=\frac{1}{9}$

$E⟶$ event of getting sum as $6$

No. of favorable outcomes $=5(i.e.,(1,5\left)\right(2,4\left)\right(3,3\left)\right(4,2\left)\right(5,1\left)\right)$

$P\left(E\right)=\frac{5}{36}$

$E⟶$ event of getting sum as $7$

No. of favorable outcomes $=6(i.e.,(1,6\left)\right(2,5\left)\right(3,4\left)\right(4,3\left)\right(5,2\left)\right(6,1\left)\right)$

$P\left(E\right)=\frac{6}{36}=\frac{1}{6}$

$E⟶$ event of getting sum as $8$

No. of favorable outcomes $=5(i.e.,(2,6\left)\right(3,5\left)\right(4,4\left)\right(5,3\left)\right(6,2\left)\right)$

$P\left(E\right)=\frac{5}{36}$

$E⟶$ event of getting sum as $9$

No. of favorable outcomes $=4(i.e.,(3,6\left)\right(4,5\left)\right(5,4\left)\right(6,3\left)\right)$

$P\left(E\right)=\frac{4}{36}=\frac{1}{9}$

$E⟶$ event of getting sum as $10$

No. of favorable outcomes $=3(i.e.,(4,6\left)\right(5,5\left)\right(6,4\left)\right)$

$P\left(E\right)=\frac{3}{36}=\frac{1}{12}$

$E⟶$ event of getting sum as $11$

No. of favorable outcomes $=2(i.e.,(6,5\left)\right(5,6\left)\right)$

$P\left(E\right)=\frac{2}{36}=\frac{1}{18}$

$E⟶$ event of getting sum as $12$

No. of favorable outcomes $=1(i.e.,(6,6\left)\right)$

$P\left(E\right)=\frac{1}{36}$

From the figure(table) we can see that the outcomes are not equally likely - we see that, there is different probability for different outcome.

Hence cannot agree with the argument.