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 $=P(r \geq 6)$
$=P(r=6)+P(r=7)+P(r=8)+P(r=9)+P(r=10)$
$={ }^{10} C_{6}\left(\frac{1}{2}\right)^{10}+{ }^{10} C_{7}\left(\frac{1}{2}\right)^{10}+{ }^{10} C_{8}\left(\frac{1}{2}\right)^{10}+{ }^{10} C_{9}\left(\frac{1}{2}\right)^{10}+\left(\frac{1}{2}\right)^{10}$
$=\frac{1}{2^{10}}\left({ }^{10} C_{6}+{ }^{10} C_{7}+{ }^{10} C_{8}+{ }^{10} C_{9}+1\right)$
$=\frac{1}{2^{10}}(210+120+45+10+1) $
$=\frac{386}{2^{10}}=\frac{193}{2^{9}}$