Question

A die is thrown twice. Find the probability of $4$ turning up at least once.

Answer 1

Finding the probability of $4$ turning up at least once.

Let $\text{P}\left(\text{E}\right)$ be the probability of $4$ turning up at least once.

All possible outcomes when a die is thrown twice$=$$$\begin{gathered} (1,1)(1,2)(1,3)(1,4)(1,5)(1,6) \\ (2,1)(2,2)(2,3)(2,4)(2,5)(2,6) \\ (3,1)(3,2)(3,3)(3,4)(3,5)(3,6) \\ (4,1)(4,2)(4,3)(4,4)(4,5)(4,6) \\ (5,1)(5,2)(5,3)(5,4)(5,5)(5,6) \\ (6,1)(6,2)(6,3)(6,4)(6,5)(6,6) \end{gathered}$$

$\therefore$Number of possible outcomes$=36$

Outcomes with $4$ turning up at least once$=\left(1,4\right),\left(2,4\right),\left(3,4\right),\left(4,1\right),\left(4,2\right),\left(4,3\right),\left(4,4\right),\left(4,5\right),\left(4,6\right),\left(5,4\right)$ and $\left(6,4\right)$

$\therefore$Number of favourable outcomes$=11$

We know that $\text{P}\left(\text{E}\right)$$=\frac{\text{Number of favourable outcomes}}{\text{Number of possible outcomes}}$

So, $\text{P}\left(\text{E}\right)$$=\frac{11}{36}$

Hence, probability of $4$ turning up at least once$=\frac{\mathbf{11}}{\mathbf{36}}$