Question

Probability of guessing correctly atleast $7$ out of $10$ answers in a "True" or "False" test is = ______

• A. $\frac{11}{64}$
• B. $\frac{11}{32}$
• C. $\frac{11}{16}$
• D. $\frac{27}{32}$

Answer 1

Answer: (A)

In each true and false question, probability
of guessing correctly is, $p=\frac{1}{2}$ and probability of
not guessing correctly, $q=\frac{1}{2}$.
Here, $n=10$
$\therefore$ The probability of guessing atleast 7 correctly
$=P(X\ge 7)$
$=P(X=7)+P(X=8)+P(X=9)+P(X=10)$
$={}^{10}{C}_7{\left(\frac{1}{2}\right)}^7{\left(\frac{1}{2}\right)}^3+{}^{10}{C}_8{\left(\frac{1}{2}\right)}^8{\left(\frac{1}{2}\right)}^2$
$+{}^{10}{C}_9{\left(\frac{1}{2}\right)}^9{\left(\frac{1}{2}\right)}^1+{}^{10}{C}_{10}{\left(\frac{1}{2}\right)}^{10}$
$\left[\because P(x=r)={}^n{C}_r{p}^{\prime }{q}^{n-r}\right]$
$=120\times {\left(\frac{1}{2}\right)}^{10}+45{\left(\frac{1}{2}\right)}^{10}+10{\left(\frac{1}{2}\right)}^{10}+1{\left(\frac{1}{2}\right)}^{10}$
$=\frac{120+45+10+1}{2^{10}}=\frac{176}{1024}=\frac{11}{64}$