Question

If a fair coin is tossed 6 times, find the probability of getting at least 4 heads.

Hint:

When the question says “at least $r$,” sum binomial probabilities from $r$ to $n$ using: \[ P(X \geq r) = \sum_{k=r}^{n} {n \choose k} p^k (1 - p)^{n - k} \]

Answer 1

This is a binomial probability problem involving coin tosses. A "success" is defined as getting a head. The number of trials is $n = 6$, and the probability of success (getting a head) is $p = \dfrac{1}{2}$.
The objective is to determine the probability of obtaining at least 4 heads, denoted as $P(X \geq 4)$.
This probability is calculated as the sum of probabilities for getting exactly 4, 5, or 6 heads:
\[P(X \geq 4) = P(X = 4) + P(X = 5) + P(X = 6)\]
The binomial probability formula is applied:
\[P(X = r) = {n \choose r} p^r (1 - p)^{n - r}\]
Calculating each component:
$P(X = 4) = {6 \choose 4} \left(\dfrac{1}{2}\right)^6 = 15 \times \dfrac{1}{64} = \dfrac{15}{64}$
$P(X = 5) = {6 \choose 5} \left(\dfrac{1}{2}\right)^6 = 6 \times \dfrac{1}{64} = \dfrac{6}{64}$
$P(X = 6) = {6 \choose 6} \left(\dfrac{1}{2}\right)^6 = 1 \times \dfrac{1}{64} = \dfrac{1}{64}$
Summing these probabilities yields:
\[P(X \geq 4) = \dfrac{15}{64} + \dfrac{6}{64} + \dfrac{1}{64} = \dfrac{15 + 6 + 1}{64} = \dfrac{22}{64}\]
This fraction simplifies to $\dfrac{11}{32}$.
Final Answer: $\boxed{\dfrac{11}{32}}$