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Q.
Eight coins are tossed together. The probability of getting exactly $3$ heads is
Probability - Part 2
Solution:
Throwing of eight coins simultaneously is the same as throwing of one coin $8$ times.
$\therefore $ Probability of getting a head, $p = \frac{1}{2}$
$\Rightarrow q = 1-p=1-\frac{1}{2} = \frac{1}{2}$
Thus, we have a binomial distribution with $p = \frac{1}{2}$, $q=\frac{1}{2}$ and $n = 8$.
Hence, required probability $=\,{}^{8}C_{3}p^{3}q^{5}$
$= 56\times\left(\frac{1}{2}\right)^{3}\left(\frac{1}{2}\right)^{5}$
$= 56\times\frac{1}{8}\times\frac{1}{32} = \frac{7}{32}$