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Q.
Eight coins are thrown simultaneously. Find the probability of getting atleast $6$ heads.
Probability - Part 2
Solution:
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 $=P\left(6\right) + P\left(7\right) + P\left(8\right)$
$= \,{}^{8}C_{6}\left(\frac{1}{2}\right)^{6}\left(\frac{1}{2}\right)^{2}+\,{}^{8}C_{7} \left(\frac{1}{2}\right)^{7}\left(\frac{1}{2}\right)+\,{}^{8}C_{8}\left(\frac{1}{2}\right)^{8}$
$= \left(^{8}C_{6}+\,{}^{8}C_{7}+\,{}^{8}C_{8}\right)\left(\frac{1}{2}\right)^{8}$
$= \left(28+8+1\right)\times\frac{1}{256} = \frac{37}{256}$