Question

Question 24 (i)
A die is thrown twice. What is the probability that 5 will not come up either time?

Answer 1

The outcomes associated with the experiment in which a die is thrown is twice:
(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 outcomes = 36
Now consider the following events:
A = First throw shows 5 and B = Second throw shows 5
Therefore, the number of favourable outcomes = 6 in each case.
$\therefore$ P(A) = $\frac{6}{36}$ and P(B) =$\frac{6}{36}$
$\Rightarrow P\left(\overline{A}\right)=1-\frac{6}{36}=\frac{30}{36}=\frac{5}{6}$ and $P\left(\overline{B}\right)=\frac{5}{6}$
$\therefore$ Required probability = $\frac{5}{6}\times \frac{5}{6}=\frac{25}{36}$