Question

Six coins are tossed simultaneously. The probability of getting at least $4$ heads is

• A. $\frac{11}{64}$
• B. $\frac{11}{32}$
• C. $\frac{15}{44}$
• D. $\frac{21}{32}$

Answer 1

Answer: (B)

$\frac{11}{32}$

$Since,probailityofgettingaheadis\frac{1}{2},therequiredprobabilityis{}^{}{}^6{C}_4{\left(\frac{1}{2}\right)}^4\times {\left(\frac{1}{2}\right)}^2{+}^6{C}_5{\left(\frac{1}{2}\right)}^5\times \left(\frac{1}{2}\right){+}^6{C}_6{\left(\frac{1}{2}\right)}^6=\left(15+6+1\right)\times {\left(\frac{1}{2}\right)}^6=\frac{22}{64}=\frac{11}{32}$

Hence, option $B$ is the correct answer.