Eight coins are thrown simultaneously. What is the probability of getting atleast 3 heads?

$\frac{219}{256}$

The correct answer is Option (2) → $\frac{219}{256}$

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 = $p =\frac{1}{2}$, so $q = 1-\frac{1}{2}=\frac{1}{2}$.

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 $n = 8$.

$P(r) = {^nC}_rp^rq^{n-r} = {^8C}_r(\frac{1}{2})^r (\frac{1}{2})^{8-r} = {^8C}_r(\frac{1}{2})^8$

Probability of getting atleast 3 heads = $P(r ≥ 3)$

$= P(3) + P(4) + P(5) + P(6) + P(7) + P(8)$

$= 1 − (P(0) + P(1) + P(2)) = 1 − ({^8C}_0 + {^8C}_1 + {^8C}_2)(\frac{1}{8})^8$

$=1-(1+8+28) ×\frac{1}{256}=1-\frac{37}{256}=\frac{219}{256}$