Question

Question: What is the probability that a positive integer not exceeding $\mathbf{100}$selected at random is divisible by $\mathbf{5}\mathbf{}\mathbf{\text{or}}\mathbf{}\mathbf{7}$?

Short answer

Answer

The probability that a positive integer not exceeding $100$selected at random is divisible by $\text{5or7isP(E)=}\frac{8}{25}$.

Step-by-step solution

 Given  

A positive integer not exceeding $100$.

The Concept of Probability

If$\mathit{S}$represents the sample space and$\mathit{E}$represents the event. Then the probability of occurrence of favourable event is given by the formula as below:

$\mathit{P}\mathbf{\left(}\mathit{E}\mathbf{\right)}\mathbf{=}\frac{\mathbf{n}\mathbf{\left(}\mathbf{E}\mathbf{\right)}}{\mathbf{n}\mathbf{\left(}\mathbf{S}\mathbf{\right)}}$.

Determine the probability

As per the problem we have been asked to find the probability that a fair die never comes up an even number when it is rolled six times.

If $S$represents the sample space and $E$represents the event. Then the probability of occurrence of favourable event is given by the formula as below:

$P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}$.

Sample space of numbers not exceeding $100\text{is}S=\{1,2,3,\dots ,100\}$

The favourable event of divisible by $5\text{is}E=\{5,10,15,\dots ,100\},n\left(E\right)=20$

The favourable event of divisible by$7\text{is}E=\{7,14,21,\dots ,98\},n\left(E\right)=14$

$\left(E_1\cap E_2\right)=\{35,70\},n\left(E_1\cap E_2\right)=2$

The event representing the divisibility by$\text{5or7is}\left(E_1\cup E_2\right)$ .

$$\begin{gathered} P\left(E_1\cup E_2\right)=P\left(E_1\right)+P\left(E_2\right)-P\left(E_1\cap E_2\right) \\ P\left(E_1\cup E_2\right)=\frac{20}{100}+\frac{14}{100}-\frac{2}{100} \\ P\left(E_1\cup E_2\right)=\frac{32}{100} \\ P\left(E_1\cup E_2\right)=\frac{8}{25} \end{gathered}$$

The probability that a positive integer not exceeding $100$selected at random is divisible by $\text{5or7isP(E)=}\frac{8}{25}$.