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)

$\frac{37}{256}$

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 $6$ heads $=P\left(6\right) + P\left(7\right) + P\left(8\right)$ $= \,^{8}C_{6}\left(\frac{1}{2}\right)^{6}\left(\frac{1}{2}\right)^{2}+\,^{8}C_{7} \left(\frac{1}{2}\right)^{7}\left(\frac{1}{2}\right)+\,^{8}C_{8}\left(\frac{1}{2}\right)^{8}$ $= \left(^{8}C_{6}+\,^{8}C_{7}+\,^{8}C_{8}\right)\left(\frac{1}{2}\right)^{8}$ $= \left(28+8+1\right)\times\frac{1}{256} = \frac{37}{256}$