Question

Four numbers are chosen at random from $\{1,2,3,......,40\}$.

The probability that they are not consecutive, is

• A. $\frac{1}{2470}$
• B. $\frac{4}{7969}$
• C. $\frac{2469}{2470}$
• D. $\frac{7965}{7969}$

Answer 1

The correct option is C

$\frac{2469}{2470}$

The explanation for the correct option:

Step:1 Calculate the total number of cases that four numbers are consecutive

The total numbers are $40$ and the numbers to be selected at random is $4$.

Therefore, $4$numbers can be chosen out of $40$ numbers in ${}^{40}C_4$different ways.

Again, $4$numbers can be chosen out of $40$ numbers at random and they are consecutive is $37$.

Therefore, the probability of the selected $4$numbers being consecutive is $\frac{37}{{}^{40}C_4}$

$$\begin{gathered} =\frac{37}{{\displaystyle \left(\frac{40!}{4!\times 36!}\right)}}\mathbf{[}\mathbf{\because }{}^{\mathbf{n}}\mathbf{C}_{\mathbf{r}}\mathbf{=}\frac{\mathbf{n}\mathbf{!}}{\mathbf{r}\mathbf{!}\mathbf{}\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{r}\mathbf{)}\mathbf{!}}\mathbf{]}\mathbf{} \\ =\frac{37\times 36!\times 4!}{40!} \\ =\frac{37!\times 4\times 3\times 2}{40\times 39\times 38\times 37!} \\ =\frac{1}{2470} \end{gathered}$$

Step:2 Calculate the total number of cases that four numbers that are not consecutive:

Now, the probability of the selected $4$numbers are not consecutive is

$$\begin{gathered} =1-\frac{1}{2470}\mathbf{\left[}\mathbf{\because }\mathbf{P}\mathbf{\left(}\overline{\mathbf{E}}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{-}\mathbf{P}\mathbf{\left(}\mathbf{E}\mathbf{\right)}\mathbf{\right]} \\ =\frac{2469}{2470} \end{gathered}$$

Hence, Option(C) is the correct answer.