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)

$\frac{219}{256}$

Answer 2

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