Question

# A die is thrown twice. What is the probability that $5$ will not come up either time?

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Solution

## Probability of an event $=\frac{numberoffavorableoutcomes}{totalnumberofoutcomes}$We know that,If a die is rolled twice, the total number of outcomes is $36$.Sample space is defined by $\left(\mathrm{S}\right)$$\mathrm{S}=\mathrm{\left\{}\left(1,1\right)\left(1,2\right)\left(1,3\right)\left(1,4\right)\left(1,5\right)\left(1,6\right)\phantom{\rule{0ex}{0ex}}\left(2,1\right)\left(2,2\right)\left(2,3\right)\left(2,4\right)\left(2,5\right)\left(2,6\right)\phantom{\rule{0ex}{0ex}}\left(3,1\right)\left(3,2\right)\left(3,3\right)\left(3,4\right)\left(3,5\right)\left(3,6\right)\phantom{\rule{0ex}{0ex}}\left(4,1\right)\left(4,2\right)\left(4,3\right)\left(4,4\right)\left(4,5\right)\left(4,6\right)\phantom{\rule{0ex}{0ex}}\left(5,1\right)\left(5,2\right)\left(5,3\right)\left(5,4\right)\left(5,5\right)\left(5,6\right)\phantom{\rule{0ex}{0ex}}\left(6,1\right)\left(6,2\right)\left(6,3\right)\left(6,4\right)\left(6,5\right)\left(6,6\right)\right\}$So, $\mathrm{n}\left(\mathrm{S}\right)=36$Step 1: Calculating the probability of getting $5$ on either timeLet $\mathrm{A}$ be the event of getting $5$ either time.The outcomes where $5$ turns up ether time are $=\left\{\left(1,5\right),\left(2,5\right),\left(3,5\right),\left(4,5\right),\left(5,1\right),\left(5,2\right),\left(5,3\right),\left(5,4\right),\left(5,5\right),\left(5,6\right),\left(6,5\right)\right\}$So, $\mathrm{n}\left(\mathrm{A}\right)=11$Therefore, the probability of getting $5$ on either time, $\mathrm{P}\left(\mathrm{A}\right)=\frac{\mathrm{n}\left(\mathrm{A}\right)}{\mathrm{n}\left(\mathrm{S}\right)}$ $\mathrm{P}\left(\mathrm{A}\right)=\frac{11}{36}$Step 2: Calculating the probability of not getting $5$ on either timeSo, the probability that $5$ will not come up on either time $=1–\mathrm{P}\left(\mathrm{A}\right)$ $=1–\frac{11}{36}\phantom{\rule{0ex}{0ex}}=\frac{25}{36}$Hence, the probability that $5$ will not come up either time is $\frac{25}{36}$.

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