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 where each toss has two outcomes: head or tail.
Let getting a head be a "success". Number of trials $n = 6$, probability of success $p = \dfrac{1}{2}$.
We are asked to find the probability of getting at least 4 heads, i.e., $P(X \geq 4)$.
So, we need to compute:
\[ P(X \geq 4) = P(X = 4) + P(X = 5) + P(X = 6) \]
We use the binomial formula: \[ P(X = r) = {n \choose r} p^r (1 - p)^{n - r} \]
Calculate each term:
$P(X = 4) = {6 \choose 4} \left(\dfrac{1}{2}\right)^6 = 15.\dfrac{1}{64} = \dfrac{15}{64}$
$P(X = 5) = {6 \choose 5} \left(\dfrac{1}{2}\right)^6 = 6.\dfrac{1}{64} = \dfrac{6}{64}$
$P(X = 6) = {6 \choose 6} \left(\dfrac{1}{2}\right)^6 = 1.\dfrac{1}{64} = \dfrac{1}{64}$
Add them together:
\[ P(X \geq 4) = \dfrac{15 + 6 + 1}{64} = \dfrac{22}{64} = \dfrac{11}{32} \]
Correction: miscount. Actually: $P(X = 4)$: $15/64$
$P(X = 5)$: $6/64$
$P(X = 6)$: $1/64$
Sum: $15 + 6 + 1 = 22$ $\Rightarrow$ $\dfrac{22}{64} = \dfrac{11}{32}$
Final Answer: $\boxed{\dfrac{11}{32}}$