Question

The probability of choosing at random a number that is divisible by $6$ or $8$ from among $1$ to $90$ is equal to

• A. $\frac{1}{6}$
• B. $\frac{1}{30}$
• C. $\frac{11}{80}$
• D. $\frac{23}{90}$

Answer 1

The correct option is D

$\frac{23}{90}$

Explanation for the correct option:

Solve the given expression,

Given expression,

Total set of numbers ,$S=\left\{1,2,3.......90\right\}$

Now, the favorable numbers that are divisible by $6$ are $6,12,18,24,30,36,42,48,54,60,66,72,78,84,90$

So , the total numbers of favorable cases $p=15$

Again, the favorable numbers that are divisible by $8$are $8,16,24,32,40,48,56,64,72,80,88$

Then, the total numbers of cases $q=11$

The favorable numbers that are common in both cases are$24,48,72$

So ,the total numbers of common cases $r=3$

$\therefore$The total numbers of numbers that are divisible by $6$ or $8$ are,

$$\begin{gathered} Z=p+q-r \\ \Rightarrow Z=15+11-3 \\ \Rightarrow Z=23 \end{gathered}$$

The probability of choosing at random a number that is divisible by $6$ or $8$

$$\begin{gathered} P\left(Z\right)=\frac{\mathbf{Total}\mathbf{}\mathbf{no}\mathbf{.}\mathbf{}\mathbf{of}\mathbf{}\mathbf{numbers}\mathbf{}\mathbf{which}\mathbf{}\mathbf{are}\mathbf{}\mathbf{divisible}\mathbf{}\mathbf{by}\mathbf{}\mathbf{6}\mathbf{}\mathbf{or}\mathbf{}\mathbf{8}}{\mathbf{Total}\mathbf{}\mathbf{set}\mathbf{}\mathbf{of}\mathbf{}\mathbf{numbers}} \\ \Rightarrow P\left(Z\right)=\frac{23}{90} \end{gathered}$$

Therefore, The probability of choosing at random a number that is divisible by $6$ or $8$ from among $1$ to $90$ is $\frac{23}{90}$.

Hence, option (D) is the correct option.