Question

A die is thrown three times.

Getting a $3$ or a $6$ is considered a success.

Then the probability of at least two successes is

• A. $\frac{2}{9}$
• B. $\frac{7}{27}$
• C. $\frac{1}{27}$
• D. None of these

Answer 1

The correct option is B

$\frac{7}{27}$

The explanation for the correct option:

Step1. Define events:

The number of times a die is thrown$\text{'}\mathrm{n}\text{'}=3$

Possible outcomes of throwing a die $=\left\{1,2,3,4,5,6\right\}$

The probability of getting $3or6$is success $\text{'}p\text{'}=\frac{2}{6}$

$=\frac{1}{3}$

The probability of not getting $3or6$is failure $\text{'}q\text{'}=1-\frac{1}{3}$

$=\frac{2}{3}$

Step2. Find the probability of getting at least two successes:

The probability of getting at least two successes is two or more than two

$\mathrm{P}(\mathrm{X}=2)+\mathrm{P}(\mathrm{X}=3)$

$$\begin{gathered} ={}^{3}C_2{\left(\frac{1}{3}\right)}^2{\left(\frac{2}{3}\right)}^{3-2}+{}^{3}C_3{\left(\frac{1}{3}\right)}^3{\left(\frac{2}{3}\right)}^{3-3}\mathbf{}\left[\because P\left(x\right)={}^{n}C_x{\left(p\right)}^x{\left(q\right)}^{n-x}\right] \\ =\frac{3!}{\left(3-2\right)!2!}\left(\frac{2}{27}\right)+\frac{3!}{\left(3-3\right)!3!}\left(\frac{1}{27}\right) \\ =\frac{6}{27}+\frac{1}{27} \\ =\frac{\mathbf{7}}{\mathbf{27}} \end{gathered}$$

Hence, Option(B) is the correct answer.