Question

In a 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 heads, he answers 'true'; if it falls tails, he answers 'false'. The probability that he answers at least $12$ questions correctly is:

• A. ${\left(\frac{1}{2}\right)}^{20}\left(\frac{2^{20}-{}^{20}{C}_{10}}{2}\right)$
• B. ${\left(\frac{1}{2}\right)}^{20}\left(\frac{2^{20}-{}^{20}{C}_{10}-2\cdot {}^{20}{C}_9}{2}\right)$
• C. ${\left(\frac{1}{2}\right)}^{20}\left(\frac{2^{20}-{}^{20}{C}_{10}-{}^{20}{C}_9}{2}\right)$
• D. ${\left(\frac{1}{2}\right)}^{20}\left(\frac{2^{20}-2\cdot {}^{20}{C}_9}{2}\right)$

Answer 1

The correct option is B ${\left(\frac{1}{2}\right)}^{20}\left(\frac{2^{20}-{}^{20}{C}_{10}-2\cdot {}^{20}{C}_9}{2}\right)$

Let us assume that the number of correctly answered questions out of twenty questions be $X.$
Since,'head' on the coin shows the true answer and the 'tail' on the coin shows the false answers. Thus, the repeated tosses or the correctly answered questions are Bernoulli trials.
Thus, $p=1/2\text{and}q=1-p=1-1/2=1/2$
Here, it can be clearly observed that $X$ has binomial distribution, where $n=20$ and $p=1/2$
Thus, $P(X=x){=}^n{C}_x{q}^{n-x}{p}^x,$ where $x=0,1,2,...n{=}^{20}{C}_x{\left(\frac{1}{2}\right)}^{20-x}{\left(\frac{1}{2}\right)}^x{=}^{20}{C}_x{\left(\frac{1}{2}\right)}^{20}$
Probability of at least 12 questions answered correctly $=P(X\ge 12)=P(X=12)+P(X=13)+\cdots +P(X=20))$
${=}^{20}{C}_{12}{\left(\frac{1}{2}\right)}^{20}{+}^{20}{C}_{13}{\left(\frac{1}{2}\right)}^{20}+\cdots {+}^{20}{C}_{20}{\left(\frac{1}{2}\right)}^{20}={\left(\frac{1}{2}\right)}^{20}{(}^{20}{C}_{12}{+}^{20}{C}_{13}+...{+}^{20}{C}_{20})$
$={\left(\frac{1}{2}\right)}^{20}\left(\frac{2^{20}-{}^{20}{C}_{10}-2\cdot {}^{20}{C}_9}{2}\right)$
$\{\because {}^{20}{C}_0+{}^{20}{C}_1+\cdots +{}^{20}{C}_{20}=2^{20}\Rightarrow {}^{20}{C}_{10}+2({}^{20}{C}_{11}+{}^{20}{C}_{12}+\cdots +{}^{20}{C}_{20})=2^{20}\Rightarrow {}^{20}{C}_{12}+\cdots +{}^{20}{C}_{20}=\frac{2^{20}-{}^{20}{C}_{10}-2\cdot {}^{20}{C}_9}{2}\}$