Question

A die is thrown twice. Find the probability that 5 will come up at least once.

Answer 1

Solution:

Step1:Sample space is:

Given: A dice is thrown twice

$$\begin{gathered} \left(1,1\right),\left(1,2\right),\left(1,3\right),\left(1,4\right),\left(1,5\right),\left(1,6\right) \\ \left(2,1\right),\left(2,2\right),\left(2,3\right),\left(2,4\right),\left(2,5\right),\left(2,6\right) \\ \left(3,1\right),\left(3,2\right),\left(3,3\right),\left(3,4\right),\left(3,5\right),\left(3,6\right) \\ \left(4,1\right),\left(4,2\right),\left(4,3\right),\left(4,4\right),\left(4,5\right),\left(4,6\right) \\ \left(5,1\right),\left(5,2\right),\left(5,3\right),\left(5,4\right),\left(5,5\right),\left(5,6\right) \\ \left(6,1\right),\left(6,2\right),\left(6,3\right),\left(6,4\right),\left(6,5\right),\left(6,6\right) \end{gathered}$$

Total numbers of outcomes = 36

Favorable outcomes:

Outcomes when 5 will come up at least once

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

Number of favorable outcomes:

Number of Outcomes when 5 will come at least once = 11

Step2: Find the probability:

Probability that 5 will come up at least once

$\frac{Numbersofoutcomeswhen5willcomeuponce}{Totalnumberofoutcomes}$

$=\frac{11}{36}$

Final Answer: Therefore, Probability of 5 will come at least once is$\frac{11}{36}$.