Question

Question 22 (i)
(i) Complete the following table:
$\begin{array}{cccccccccccc}\text{Event: Sum of 2 dice}& 2& 3& 4& 5& 6& 7& 8& 9& 10& 11& 12\\ \text{Probability}& \frac{1}{36}& & & & & & \frac{5}{36}& & & & \frac{1}{36}\end{array}$

Answer 1

Total favourable outcomes of throwing two dice 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)
$\therefore$ Total number of favourable outcomes = 36
(i) Favourable outcomes of getting the sum as 3 = 2
Hence, P (getting the sum as 3) = $\frac{2}{36}=\frac{1}{18}$
Favourable outcomes of getting the sum as 4 = 3
Hence, P (getting the sum as 4) = $\frac{3}{36}=\frac{1}{12}$
Favourable outcomes of getting the sum as 5 = 4
Hence, P (getting the sum as 5) = $\frac{4}{36}=\frac{1}{9}$
Favourable outcomes of getting the sum as 6 = 5
Hence, P (getting the sum as 6)=$\frac{5}{36}$
Favourable outcomes of getting the sum as 7 = 6
Hence, P (getting the sum as 7) = $\frac{6}{36}=\frac{1}{6}$
Favourable outcomes of getting the sum as 9 = 4
Hence, P (getting the sum as 9) = $\frac{4}{36}=\frac{1}{9}$
Favourable outcomes of getting the sum as 10 = 3
Hence, P (getting the sum as 10) = $\frac{3}{36}=\frac{1}{12}$
Favourable outcomes of getting the sum as 11 = 2
Hence, P (getting the sum as 11) = $\frac{2}{36}=\frac{1}{18}$
$\begin{array}{cccccccccccc}\text{Event: Sum of 2 dice}& 2& 3& 4& 5& 6& 7& 8& 9& 10& 11& 12\\ \text{Probability}& \frac{1}{36}& \frac{2}{36}& \frac{3}{36}& \frac{4}{36}& \frac{5}{36}& \frac{6}{36}& \frac{5}{36}& \frac{4}{36}& \frac{3}{36}& \frac{2}{36}& \frac{1}{36}\end{array}$