Question

Eight coins are thrown simultaneously. Find the probability of getting atleast $6$ heads.

• A. $\frac{31}{128}$
• B. $\frac{37}{256}$
• C. $\frac{37}{128}$
• D. $\frac{31}{256}$

Answer 1

Answer: (B)

Throwing of eight coins simultaneously is the same as throwing of one coin $8$ times. When a coin is thrown, the probability of getting a head, $p=\frac{1}{2}$,
$\therefore q=1-\frac{1}{2}=\frac{1}{2}$
As the coin is thrown $8$ times, so there are $8$ trials.
Thus, we have a binomial distribution with $p=\frac{1}{2}$,
$q=\frac{1}{2}$ and $n=8$.
Probability of getting atleast $6$ heads $=P\left(6\right)+P\left(7\right)+P\left(8\right)$
$={}^8{C}_6{\left(\frac{1}{2}\right)}^6{\left(\frac{1}{2}\right)}^2+{}^8{C}_7{\left(\frac{1}{2}\right)}^7\left(\frac{1}{2}\right)+{}^8{C}_8{\left(\frac{1}{2}\right)}^8$
$=\left({}^8{C}_6+{}^8{C}_7+{}^8{C}_8\right){\left(\frac{1}{2}\right)}^8$
$=\left(28+8+1\right)\times \frac{1}{256}=\frac{37}{256}$