Question

A contest consists of predicting the results (win, draw or defeat) of $10$ football matches. The probability that one entry contains at least $5$ correct answers is

• A. $\frac{12585}{3^{10}}$
• B. $\frac{12385}{3^{10}}$
• C. $\frac{9385}{3^{10}}$
• D. None of these

Answer 1

The correct option is A

$\frac{12585}{3^{10}}$

Explanation for the correct answer:

Step 1: Find the probability of success and failure for a given experiment

We have been given that, a contest consists of predicting the results (win, draw or defeat) of $10$ football matches.

We need to find the probability that one entry contains at least $5$ correct answers.

Here, the probability of predicting the correct result $p=\frac{1}{3}$

the probability of predicting the wrong result $q=\frac{2}{3}$

$n=10$

Step 2: By using the Binomial distribution formula, find the probability of predicting exactly $5$ correct answers.

$\begin{array}{rcl}& \Rightarrow & P(x=5)={}^{10}C_5\times {\left(\frac{1}{3}\right)}^5\times {\left(\frac{2}{3}\right)}^{10-5}=\frac{10!}{5!(10-5)!}\times \frac{1}{3^5}\times {\left(\frac{2}{3}\right)}^5\\ & =& 252\times \frac{1}{3^5}\times \frac{32}{3^5}\\ & =& \frac{8064}{3^5\times 3^5}\\ & =& \frac{8064}{3^{10}}\end{array}$

Step 3: By using the Binomial distribution formula, find the probability of predicting exactly $6$ correct answers.

$\begin{array}{rcl}& \Rightarrow & P(x=6)={}^{10}C_6\times {\left(\frac{1}{3}\right)}^6\times {\left(\frac{2}{3}\right)}^{10-6}=\frac{10!}{6!(10-6)!}\times \frac{1}{3^6}\times {\left(\frac{2}{3}\right)}^4\\ & =& 210\times \frac{1}{3^6}\times \frac{16}{3^4}\\ & =& \frac{3360}{3^6\times 3^4}\\ & =& \frac{3360}{3^{10}}\end{array}$

Step 4: By using the Binomial distribution formula, find the probability of predicting exactly $7$ correct answers.

$\begin{array}{rcl}& \Rightarrow & P(x=7)={}^{10}C_7\times {\left(\frac{1}{3}\right)}^7\times {\left(\frac{2}{3}\right)}^{10-7}=\frac{10!}{7!(10-7)!}\times \frac{1}{3^7}\times {\left(\frac{2}{3}\right)}^3\\ & =& 120\times \frac{1}{3^7}\times \frac{8}{3^3}\\ & =& \frac{960}{3^7\times 3^3}\\ & =& \frac{960}{3^{10}}\end{array}$

Step 5: By using the Binomial distribution formula, find the probability of predicting exactly $8$ correct answers.

$\begin{array}{rcl}& \Rightarrow & P(x=8)={}^{10}C_8\times {\left(\frac{1}{3}\right)}^8\times {\left(\frac{2}{3}\right)}^{10-8}=\frac{10!}{8!(10-8)!}\times \frac{1}{3^8}\times {\left(\frac{2}{3}\right)}^2\\ & =& 45\times \frac{1}{3^8}\times \frac{4}{3^2}\\ & =& \frac{180}{3^8\times 3^2}\\ & =& \frac{180}{3^{10}}\end{array}$

Step 6: By using the Binomial distribution formula, find the probability of predicting exactly $9$ correct answers.

$\begin{array}{rcl}& \Rightarrow & P(x=9)={}^{10}C_9\times {\left(\frac{1}{3}\right)}^9\times {\left(\frac{2}{3}\right)}^{10-9}=\frac{10!}{9!(10-9)!}\times \frac{1}{3^9}\times {\left(\frac{2}{3}\right)}^1\\ & =& 10\times \frac{1}{3^9}\times \frac{2}{3}\\ & =& \frac{20}{3^9\times 3}\\ & =& \frac{20}{3^{10}}\end{array}$

Step 7: By using the Binomial distribution formula, find the probability of predicting exactly $10$ correct answers.

$\begin{array}{rcl}& \Rightarrow & P(x=10)={}^{10}C_{10}\times {\left(\frac{1}{3}\right)}^{10}\times {\left(\frac{2}{3}\right)}^{10-10}=\frac{10!}{10!(10-10)!}\times \frac{1}{3^{10}}\times {\left(\frac{2}{3}\right)}^0\\ & =& 1\times \frac{1}{3^{10}}\times 1\\ & =& \frac{1}{3^{10}}\end{array}$

Step 8: Find the required probability that one entry contains at least $\mathbf{5}$ correct answers.

The probability that one entry contains at least $5$ correct answers would be,

$$\begin{gathered} \Rightarrow P=P(x=5)+P(x=6)+P(x=7)+P(x=8)+P(x=9)+P(x=10) \\ \Rightarrow P=\frac{8064}{3^{10}}+\frac{3360}{3^{10}}+\frac{960}{3^{10}}+\frac{180}{3^{10}}+\frac{20}{3^{10}}+\frac{1}{3^{10}} \\ \Rightarrow P=\frac{12585}{3^{10}} \end{gathered}$$

Therefore, option (A) is the correct answer.