Discussion Forum

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 

Answer:

Explanation:

n=10, $p= \frac{1}{2}$, q= $\frac {1}{2}$

$\therefore$  Required probability = $P(r\geq6)$

 $=P(r=6)+P(r=7)+P(r=8)+P(r=9)+P(r=10)$

$=^{10}C_{6}\left(\frac{1}{2}\right)^{10}+^{10}C_{7}\left(\frac{1}{2}\right)^{10}+^{10}C_{8}\left(\frac{1}{2}\right)^{10}+^{10}C_{9}\left(\frac{1}{2}\right)^{10}+\left(\frac{1}{2}\right)^{10}$

$=\frac{1}{2^{10}}\left(^{10}C_{6}+^{10}C_{7}+^{10}C_{8}+^{10}C_{9}+1\right)$

 $=\frac{1}{2^{10}}\left(210+120+45+10+1\right)=\frac{386}{2^{10}}=\frac{193}{2^{9}}$