Question

Eight coins are thrown simultaneously. What is the probability of getting atleast $3$ heads?

• A. $\frac{37}{246}$
• B. $\frac{21}{256}$
• C. $\frac{219}{256}$
• D. $\frac{19}{246}$

Answer 1

Answer: (C)

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 $3$ heads $=P\left(X\ge 3\right)$
$=1-\left(P\left(0\right)+P\left(1\right)+P\left(2\right)\right)$
$=1-\left[{}^8{C}_0+{\left(\frac{1}{2}\right)}^8+{}^8{C}_1\left(\frac{1}{2}\right){\left(\frac{1}{2}\right)}^7+{}^8{C}_2{\left(\frac{1}{2}\right)}^2{\left(\frac{1}{2}\right)}^6\right]$
$=1-\left({}^8{C}_0+{}^8{C}_1+{}^8{C}_2\right){\left(\frac{1}{2}\right)}^8$
$=1-\left(1+8+28\right)\times \frac{1}{256}$
$=1-\frac{37}{256}=\frac{219}{256}$