# A coin is tossed 5 times. Find the probability of getting (i) at least 4 heads (ii) at most 4 heads

A coin is tossed 5 times.

$\begin{array}{l} \text { (i) Probability (at least } 4 \text { heads }) \Rightarrow p(x=4)+p(x=5) \\ \Rightarrow^{5} C_{4}\left(\frac{1}{2}\right)^{1}\left(\frac{1}{2}\right)^{4}+{ }^{5} C_{5}\left(\frac{1}{2}\right)^{0}\left(\frac{1}{2}\right)^{5} \\ \Rightarrow^{5} C_{4}\left(\frac{1}{2}\right)^{5}+{ }^{5} C_{5}\left(\frac{1}{2}\right)^{5} \\ \Rightarrow \frac{6}{32}=\frac{3}{16} \\ \text { (ii) } Probability \text { (atmost } 4 \text { heads) }=p(x=0)+p(x=1)+p(x=2)+p(x=3)+p(x=4) \\ \left.\Rightarrow \frac{1}{2}\right)^{5}+{ }^{5} C_{1}\left(\frac{1}{2}\right)^{5}+{ }^{5} C_{2}\left(\frac{1}{2}\right)^{5}+{ }^{5} C_{3}\left(\frac{1}{2}\right)^{5}+{ }^{5} C_{4}\left(\frac{1}{2}\right)^{5} \\ \Rightarrow\left(\frac{1}{2}\right)^{5}[1+5+10+10+5] \\ \Rightarrow \frac{31}{32} \end{array}$