Question

If getting a head on a coin when it is tossed is considered as success, then the probability of having more number of failures when ten fair coins are tossed simultaneously, is

• A. $\frac{105}{2^8}$
• B. $\frac{73}{2^7}$
• C. $\frac{193}{2^9}$
• D. $\frac{638}{2^{10}}$

Answer 1

Answer: (C)

$n=10,p=\frac{1}{2},q=\frac{1}{2}$
$\therefore$ Required probability $=P(r\ge 6)$
$=P(r=6)+P(r=7)+P(r=8)+P(r=9)+P(r=10)$
$={}^{10}{C}_6{\left(\frac{1}{2}\right)}^{10}+{}^{10}{C}_7{\left(\frac{1}{2}\right)}^{10}+{}^{10}{C}_8{\left(\frac{1}{2}\right)}^{10}+{}^{10}{C}_9{\left(\frac{1}{2}\right)}^{10}+{\left(\frac{1}{2}\right)}^{10}$
$=\frac{1}{2^{10}}\left({}^{10}{C}_6+{}^{10}{C}_7+{}^{10}{C}_8+{}^{10}{C}_9+1\right)$
$=\frac{1}{2^{10}}(210+120+45+10+1)$
$=\frac{386}{2^{10}}=\frac{193}{2^9}$