Question

Mark the correct alternative in the following question:

The probability of guessing correctly at least 8 out of 10 answers of a true false type examination is

$\left(\mathrm{a}\right)\frac{7}{64}\left(\mathrm{b}\right)\frac{7}{128}\left(\mathrm{c}\right)\frac{45}{1024}\left(\mathrm{d}\right)\frac{7}{41}$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ p=\mathrm{probabiltiy}\mathrm{of}\mathrm{guessing}\mathrm{the}\mathrm{answer}\mathrm{of}\mathrm{a}\mathrm{true}\mathrm{false}\mathrm{correctly}=\frac{1}{2}\mathrm{and} \\ q=\mathrm{probabiltiy}\mathrm{of}\mathrm{guessing}\mathrm{the}\mathrm{answer}\mathrm{of}\mathrm{a}\mathrm{true}\mathrm{false}\mathrm{incorrectly}=1-p=1-\frac{1}{2}=\frac{1}{2} \\ \mathrm{Let}X\mathrm{denote}\mathrm{a}\mathrm{success}\mathrm{of}\mathrm{guessing}\mathrm{the}\mathrm{answer}\mathrm{correctly}.\mathrm{Then}, \\ X\mathrm{follows}\mathrm{the}\mathrm{binomial}\mathrm{distribution}\mathrm{with}\mathrm{parameters}n=10\mathrm{and}p=\frac{1}{2} \\ \therefore \mathrm{P}\left(X=r\right)={}^{10}\mathrm{C}_r{p}^r{q}^{\left(10-r\right)}={}^{10}\mathrm{C}_r{\left(\frac{1}{2}\right)}^r{\left(\frac{1}{2}\right)}^{\left(10-r\right)}={}^{10}\mathrm{C}_r{\left(\frac{1}{2}\right)}^{10}=\frac{{}^{10}\mathrm{C}_r}{2^{10}} \\ \mathrm{Now}, \\ \mathrm{Required}\mathrm{probability}=\mathrm{P}\left(X\ge 8\right) \\ =\mathrm{P}\left(X=8\right)+\mathrm{P}\left(X=9\right)+\mathrm{P}\left(X=10\right) \\ =\frac{{}^{10}\mathrm{C}_8}{2^{10}}+\frac{{}^{10}\mathrm{C}_9}{2^{10}}+\frac{{}^{10}\mathrm{C}_{10}}{2^{10}} \\ =\frac{45+10+1}{2^{10}} \\ =\frac{56}{1024} \\ =\frac{7}{128} \end{gathered}$$

Hence, the correct alternative is option (b).