Question

Six coins are tossed simultaneously the probability of getting at least $4$ heads is___________.

Answer 1

Explanation:

We know that,

$ProbabilityofeventtohappenP\left(E\right)=\frac{Numberoffavourableoutcomes}{TotalNumberofoutcomes}$

The probability of getting head is $\frac{1}{2}$,

The required probability is

$$\begin{gathered} ={}^6{C}_4{\left(\frac{1}{2}\right)}^4\times {\left(\frac{1}{2}\right)}^2{+}^6{C}_5{\left(\frac{1}{2}\right)}^5\times \left(\frac{1}{2}\right){+}^6{C}_6{\left(\frac{1}{2}\right)}^6 \\ =\left(15+6+1\right)\times {\left(\frac{1}{2}\right)}^6 \\ =\frac{22}{64} \\ =\frac{11}{32} \end{gathered}$$

Hence, the probability of getting at least $4$ heads is $\frac{11}{32}$.