Question

An ordinary dice is rolled for a certain number of times.

If the probability of getting an odd number $2$ times is equal to the probability of getting an even number $3$ times, then the probability of getting an odd number for odd number of times is:

• A. $\frac{3}{16}$
• B. $\frac{1}{2}$
• C. $\frac{5}{16}$
• D. $\frac{1}{32}$

Answer 1

The correct option is B

$\frac{1}{2}$

The explanation for the correct option:

Step1. Find out the probability of getting an odd number $2$ times if the dice are thrown $n$ times:

The dice has $6$ numbers on it and out of them $3$ are odd number.

Therefore, the probability of getting an odd number in a single throw is

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

Now, the probability of getting odd number $2$ times if the dice is thrown $n$ times is

$$\begin{gathered} ={}^{n}C_2{\left(\frac{1}{2}\right)}^2{\left(\frac{1}{2}\right)}^{(n-2)} \\ ={}^{n}C_2{\left(\frac{1}{2}\right)}^n \end{gathered}$$ $\mathbf{\left[}\mathbf{}\mathbf{\because }\mathbf{P}\mathbf{(}\mathbf{X}\mathbf{=}\mathbf{r}\mathbf{)}\mathbf{=}{}^{\mathbf{n}}\mathbf{C}_{\mathbf{r}}{\mathbf{\left(}\mathbf{p}\mathbf{\right)}}^{\mathbf{r}}{\mathbf{\left(}\mathbf{q}\mathbf{\right)}}^{\mathbf{n}\mathbf{-}\mathbf{r}}\mathbf{\right]}$

Step2: Find out the the probability of getting an even number $3$ times if the dice is thrown $n$ times:

Therefore, the probability of getting an even number in a single throw is

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

Now, the probability of getting even number $3$ times if the dice is thrown $n$ times is

$$\begin{gathered} {=}^n{C}_3{\left(\frac{1}{2}\right)}^3{\left(\frac{1}{2}\right)}^{(n-3)} \\ {=}^n{C}_3{\left(\frac{1}{2}\right)}^n \end{gathered}$$

Step: 3 Calculate the value of $n$.

Given that the probability of getting an odd number $2$ times is equal to the probability of getting an even number $3$ times

$\Rightarrow$ ${}^{n}C_2{\left(\frac{1}{2}\right)}^n{=}^n{C}_3{\left(\frac{1}{2}\right)}^n$

$\Rightarrow$ $\frac{n!}{2!(n-2)!}=\frac{n!}{3!(n-3)!}$

$\Rightarrow$ $\frac{1}{(n-2)!}=\frac{1}{3(n-3)!}$

$\Rightarrow$$\frac{1}{(n-2)(n-3)!}=\frac{1}{3(n-3)!}$

$\Rightarrow$ $(n-2)=3$

$\Rightarrow$ $n=5$

Step: 4 Calculate the probability getting an odd number for an odd number of times:

The required probability is $P\left(1\right)+P\left(3\right)+P\left(5\right)$ where,

$P\left(1\right)$ is the probability of getting $1$ on the dice

$P\left(3\right)$ is the probability of getting $3$ on the dice

$P\left(5\right)$ is the probability of getting $5$ on the dice

$\begin{array}{rcl}\therefore P\left(1\right)+P\left(3\right)+P\left(5\right)& =& {}^{5}C_1{\left(\frac{1}{2}\right)}^5+{}^{5}C_3{\left(\frac{1}{2}\right)}^5+{}^{5}C_5{\left(\frac{1}{2}\right)}^5\\ & =& {\left(\frac{1}{2}\right)}^5\left(5+\frac{5!}{3!\times 2!}+1\right)\\ & =& {\left(\frac{1}{2}\right)}^5\left(6+\frac{5\times 4\times 3!}{3!\times 2!}\right)\\ & =& {\left(\frac{1}{2}\right)}^5[6+10]\\ & =& {\left(\frac{1}{2}\right)}^5\times 2^4\\ & =& \frac{\mathbf{1}}{\mathbf{2}}\end{array}$

Hence, Option(B) is the correct answer.