Question

Question: What is the probability that a positive integer not exceeding $\mathbf{100}$selected at random is divisible by $\mathbf{3}$?

Short answer

Answer

The probability that a positive integer not exceeding 100 selected at random is divisible by $3$is $P\left(E\right)=0.33$.

Step-by-step solution

 Given  

A pack of cards. In a pack of cards there are 52 cards.

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 $100$is $S=\{1,2,3,\dots ,100\}$

The favourable event is$E=\{3,6,9,\dots ,99\}$

So, we have

$$\begin{gathered} n\left(s\right)=100 \\ n\left(E\right)=\frac{100-1}{3} \\ n\left(E\right)=\frac{99}{3} \\ =33 \end{gathered}$$

Substitute the values we get

$$\begin{gathered} P\left(E\right)=\frac{33}{100} \\ P\left(E\right)=0.33 \end{gathered}$$

The probability that a positive integer not exceeding $100$selected at random is divisible by 3 is $P\left(E\right)=0.33$.