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Q.
Eight coins are thrown simultaneously. What is the probability of getting atleast $3$ 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 $3$ heads $= P\left(X \ge 3\right)$
$= 1-\left(P\left(0\right) + P\left(1\right) + P\left(2\right)\right)$
$=1-\left[\,{}^{8}C_{0}+\left(\frac{1}{2}\right)^{8}+\,{}^{8}C_{1}\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)^{7}+\,{}^{8}C_{2}\left(\frac{1}{2}\right)^{2}\left(\frac{1}{2}\right)^{6}\right]$
$= 1-\left(^{8}C_{0}+\,{}^{8}C_{1}+\,{}^{8}C_{2}\right)\left(\frac{1}{2}\right)^{8}$
$= 1 - \left( 1 + 8 + 28\right)\times\frac{1}{256}$
$= 1-\frac{37}{256} = \frac{219}{256}$