Question

8 coins are tossed simultaneously. Find the probability of getting at least 6 heads.

Answer 1

The required probility${=}^8{C}_6{\left(\frac{1}{2}\right)}^6{\left(\frac{1}{2}\right)}^2{+}^8{C}_7{\left(\frac{1}{2}\right)}^7\frac{1}{2}{+}^8{C}_8{\left(\frac{1}{2}\right)}^8$

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

$=(\frac{8\times 7\times 6!}{6!2!}+\frac{8!}{7!1!}+1){\left(\frac{1}{2}\right)}^8$

$=(28+8+1){\left(\frac{1}{2}\right)}^8$

$=\frac{37}{2^8}=\frac{37}{256}$