Question

A dice is thrown $(2n+1)$ times. The probability of getting $1,3,$or $4$ at most $n$ times is

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. None of these

Answer 1

The correct option is A

$\frac{1}{2}$

Explanation for the correct option:

Step 1: Find the probability of success and failure:

We have been given that, a dice is thrown $(2n+1)$ times.

We need to find the probability of getting $1,3,$or $4$ at most $n$ times.

the probability of getting $1,3,$or $4$ would be,

$$\begin{gathered} \Rightarrow p=\frac{3}{6} \\ \Rightarrow p=\frac{1}{2} \end{gathered}$$

the probability of not getting $1,3,$or $4$ would be,

$$\begin{gathered} \Rightarrow q=\frac{3}{6} \\ \Rightarrow q=\frac{1}{2} \end{gathered}$$

Step 2: Using the Binomial distribution, find the required probability:

The probability of getting 1, 3, or 4 at most $n$ times is:

$P={}^{2n+1}C_0{\left(p\right)}^0{\left(q\right)}^{(2n+1)-0}+{}^{2n+1}C_1{\left(p\right)}^1{\left(q\right)}^{(2n+1)-1}+...+{}^{2n+1}C_n{\left(p\right)}^n{\left(q\right)}^{(2n+1)-\left(n\right)}$

Now putting the value of $p$and $q$we get

$$\begin{gathered} P={}^{2n+1}C_0{\left(\frac{1}{2}\right)}^0{\left(\frac{1}{2}\right)}^{(2n+1)-0}+{}^{2n+1}C_1{\left(\frac{1}{2}\right)}^1{\left(\frac{1}{2}\right)}^{(2n+1)-1}+...+{}^{2n+1}C_{2n+1}{\left(\frac{1}{2}\right)}^{2n+1}{\left(\frac{1}{2}\right)}^{(2n+1)-(2n+1)} \\ P={\left(\frac{1}{2}\right)}^{(2n+1)}\left({}^{2n+1}C_0+{}^{2n+1}C_1+............+{}^{2n+1}C_n\right) \\ P=\frac{1}{2^{(2n+1)}}\times {\left(2\right)}^{2n} \\ P=\frac{1}{2} \end{gathered}$$

Hence, option (A) is the correct answer.