Question

In an examination, 20 questions of true-false type are asked. Suppose a student tosses a fair coin to determine his answer to each question. If the coin falls head, he answers true, if falls tails, he answers false. Find the probability that he answers at least 12 questions correctly.

Answer 1

Hint: To solve the question given above, we will use binomial distribution of probability. For this, we will first determine the total number of trails, the probability of success and the probability of failure respectively. Then we will put all these values in the formula of binomial distribution of probability as shown below: $p\left( x=R \right){{=}^{n}}{{C}_{r}}{{p}^{r}}{{q}^{n-r}}$
Where p is the probability of success, q is the probability of failure and n is the total number of trails.

Complete step-by-step answer :
To solve this question, we will use the method of Binomial distribution of probability. According to this method, the probability of getting exactly r success in n trials is given by the following formula: $p\left( x=R \right){{=}^{n}}{{C}_{r}}{{p}^{r}}{{q}^{n-r}}$
where, p= probability of success, q= probability of failure. So to use this formula, we will first determine the total number of trails. There are 20 true-false questions and the student will answer all of them, So the total number of trials is 20. Now it is given that he will answer true if outcome of coin tossing is heads and false if outcome of coin tossing is false. Thus, the probability that the answer will be correct is $\dfrac{1}{2}$ because there are two options available and one of them is correct. Thus the probability of failure will be =$\dfrac{1}{2}$. Now, the students have to answer at least 12 questions correctly. The total number of questions he can answer are: 12, 13, 14, 15, 16, 17, 18, 19 and 20. So, the total probability will be as shown:
Total probability$\begin{aligned} & \\\ & =p\left( x=12 \right)+p\left( x=13 \right)+p\left( x=14 \right)+p\left( x=15 \right)+ \\\ & p\left( x=16 \right)+p\left( x=17 \right)+p\left( x=18 \right)+p\left( x=19 \right)+p\left( x=20 \right) \\\ \end{aligned}$
Now, we know that, p=$\dfrac{1}{2}$, q=$\dfrac{1}{2}$, n=20. So we will get:
Total probability $=\left[ ^{20}{{C}_{12}}{{\left( \dfrac{1}{2} \right)}^{12}}{{\left( \dfrac{1}{2} \right)}^{20-12}} \right]+\left[ ^{20}{{C}_{13}}{{\left( \dfrac{1}{2} \right)}^{13}}{{\left( \dfrac{1}{2} \right)}^{20-13}} \right]+\left[ ^{20}{{C}_{14}}{{\left( \dfrac{1}{2} \right)}^{14}}{{\left( \dfrac{1}{2} \right)}^{20-14}} \right]$
$\begin{aligned} & +\left[ ^{20}{{C}_{15}}{{\left( \dfrac{1}{2} \right)}^{15}}{{\left( \dfrac{1}{2} \right)}^{20-15}} \right]+\left[ ^{20}{{C}_{16}}{{\left( \dfrac{1}{2} \right)}^{16}}{{\left( \dfrac{1}{2} \right)}^{20-16}} \right]+\left[ ^{20}{{C}_{17}}{{\left( \dfrac{1}{2} \right)}^{17}}{{\left( \dfrac{1}{2} \right)}^{20-17}} \right] \\\ & +\left[ ^{20}{{C}_{18}}{{\left( \dfrac{1}{2} \right)}^{18}}{{\left( \dfrac{1}{2} \right)}^{20-18}} \right]+\left[ ^{20}{{C}_{19}}{{\left( \dfrac{1}{2} \right)}^{19}}{{\left( \dfrac{1}{2} \right)}^{20-19}} \right]+\left[ ^{20}{{C}_{20}}{{\left( \dfrac{1}{2} \right)}^{20}}{{\left( \dfrac{1}{2} \right)}^{20-20}} \right] \\\ \end{aligned}$
Total probability $=\left[ ^{20}{{C}_{12}}{{\left( \dfrac{1}{2} \right)}^{12}}{{\left( \dfrac{1}{2} \right)}^{8}} \right]+\left[ ^{20}{{C}_{13}}{{\left( \dfrac{1}{2} \right)}^{13}}{{\left( \dfrac{1}{2} \right)}^{7}} \right]+\left[ ^{20}{{C}_{14}}{{\left( \dfrac{1}{2} \right)}^{14}}{{\left( \dfrac{1}{2} \right)}^{6}} \right]$
$\begin{aligned} & +\left[ ^{20}{{C}_{15}}{{\left( \dfrac{1}{2} \right)}^{15}}{{\left( \dfrac{1}{2} \right)}^{5}} \right]+\left[ ^{20}{{C}_{16}}{{\left( \dfrac{1}{2} \right)}^{16}}{{\left( \dfrac{1}{2} \right)}^{4}} \right]+\left[ ^{20}{{C}_{17}}{{\left( \dfrac{1}{2} \right)}^{17}}{{\left( \dfrac{1}{2} \right)}^{3}} \right] \\\ & +\left[ ^{20}{{C}_{18}}{{\left( \dfrac{1}{2} \right)}^{18}}{{\left( \dfrac{1}{2} \right)}^{2}} \right]+\left[ ^{20}{{C}_{19}}{{\left( \dfrac{1}{2} \right)}^{19}}{{\left( \dfrac{1}{2} \right)}^{1}} \right]+\left[ ^{20}{{C}_{20}}{{\left( \dfrac{1}{2} \right)}^{20}}{{\left( \dfrac{1}{2} \right)}^{0}} \right] \\\ \end{aligned}$
Total probability $=\left[ ^{20}{{C}_{12}}{{\left( \dfrac{1}{2} \right)}^{20}} \right]+\left[ ^{20}{{C}_{13}}{{\left( \dfrac{1}{2} \right)}^{20}} \right]+\left[ ^{20}{{C}_{14}}{{\left( \dfrac{1}{2} \right)}^{20}} \right]$
$\begin{aligned} & +\left[ ^{20}{{C}_{15}}{{\left( \dfrac{1}{2} \right)}^{20}} \right]+\left[ ^{20}{{C}_{16}}{{\left( \dfrac{1}{2} \right)}^{20}} \right]+\left[ ^{20}{{C}_{17}}{{\left( \dfrac{1}{2} \right)}^{20}} \right] \\\ & +\left[ ^{20}{{C}_{18}}{{\left( \dfrac{1}{2} \right)}^{20}} \right]+\left[ ^{20}{{C}_{19}}{{\left( \dfrac{1}{2} \right)}^{20}} \right]+\left[ ^{20}{{C}_{20}}{{\left( \dfrac{1}{2} \right)}^{20}} \right] \\\ \end{aligned}$
Total probability $={{\left( \dfrac{1}{2} \right)}^{20}}\left[ ^{20}{{C}_{12}}{{+}^{20}}{{C}_{13}}{{+}^{20}}{{C}_{14}}{{+}^{20}}{{C}_{15}}{{+}^{20}}{{C}_{16}}{{+}^{20}}{{C}_{17}}{{+}^{20}}{{C}_{18}}{{+}^{20}}{{C}_{19}}{{+}^{20}}{{C}_{20}} \right]$
Now, we can also write $^{x}{{C}_{y}}$ as $^{x}{{C}_{x-y}}.$ So, we will get:
Total probability $={{\left( \dfrac{1}{2} \right)}^{20}}\left[ ^{20}{{C}_{8}}{{+}^{20}}{{C}_{7}}{{+}^{20}}{{C}_{6}}{{+}^{20}}{{C}_{5}}{{+}^{20}}{{C}_{4}}{{+}^{20}}{{C}_{3}}{{+}^{20}}{{C}_{2}}{{+}^{20}}{{C}_{1}}{{+}^{20}}{{C}_{0}} \right]$
Now, we will use the formula given below for calculating $^{n}{{C}_{r}}$
$^{n}{{C}_{r}}=\dfrac{n\left( n-1 \right)\left( n-2 \right).............\left( n-r+1 \right)}{1.2.3.4............r}$
Thus, $^{20}{{C}_{8}}=\dfrac{20\times 19\times 18\times 17\times 16\times 15\times 14\times 13}{1\times 2\times 3\times 4\times 5\times 6\times 7\times 8}=125970$
$\begin{aligned} & ^{20}{{C}_{7}}=\dfrac{20\times 19\times 18\times 17\times 16\times 15\times 14}{1\times 2\times 3\times 4\times 5\times 6\times 7}=77520 \\\ & ^{20}{{C}_{6}}=\dfrac{20\times 19\times 18\times 17\times 16\times 15}{1\times 2\times 3\times 4\times 5\times 6}=38760 \\\ & ^{20}{{C}_{5}}=\dfrac{20\times 19\times 18\times 17\times 16}{1\times 2\times 3\times 4\times 5}=15504 \\\ & ^{20}{{C}_{4}}=\dfrac{20\times 19\times 18\times 17}{1\times 2\times 3\times 4}=4845 \\\ & ^{20}{{C}_{3}}=\dfrac{20\times 19\times 18}{1\times 2\times 3}=1140 \\\ & ^{20}{{C}_{2}}=\dfrac{20\times 19}{1\times 2}=190 \\\ & ^{20}{{C}_{1}}=\dfrac{20}{1}=20 \\\ & ^{20}{{C}_{0}}=1 \\\ \end{aligned}$
Thus, we will put all these values in the total probability. Thus, we will get:
Total probability $=\dfrac{125970+77520+38760+15504+4845+1140+190+20+1}{{{\left( 2 \right)}^{20}}}$
Total probability $=\dfrac{263950}{{{2}^{20}}}=\dfrac{131975}{{{2}^{19}}}$

Note: We can also apply binomial distribution of probability in the following way. First we will find the probabilities of the cases when the student will answer less than 12 questions correctly. Then we will add all these cases and we will subtract it from 1. This is shown below:
Total probability $=1-\left[ p\left( x=0 \right)+p\left( x=1 \right)+p\left( x=2 \right)+............+p\left( x=10 \right)+p\left( x=11 \right) \right]$
Where $p\left( x=r \right){{=}^{n}}{{C}_{r}}{{p}^{r}}{{q}^{n-r}}.$