If five coins are flipped simultaneously, find the probability of each of the following: 
If five coins are flipped simultaneously, find the probability of each of the following:
(a) At least one coin lands heads;

My answer:$\frac{2^5 -1}{2^5}$. I took the complement for the numerator.

(b) At most one coin lands heads.

My answer: $\frac{6}{2^5}$. I counted how many times heads appears $1$ or $0$ times.
Is this correct?
 A: Good job!!
For part a, notice that this is the OPPOSITE of (no coin lands heads <=> all coins land tails), which is $\displaystyle \left(\frac12 \right)^5=\frac{1}{32}$ chance.
For part b, any $5$ of the coins can land heads, if in total one coin lands heads. If no coin lands heads, that can be done in $1$ way. That is $\displaystyle \frac{6}{32}=\frac{3}{16}$ chance.
In general, take a look at the binomial distribution.
A: Your answers are correct. In terms of notation I would use the binomial distribution, where the probability of $k$ successes out of $n$ attempts equals:
$${n \choose k} p^k (1-p)^{n-k}$$


*

*The probability of at least one heads equals 1 minus the probability of no heads:
$$1 - {5 \choose 0} \bigg(\frac{1}{2}\bigg)^0 \bigg(\frac{1}{2}\bigg)^5 = 1 - \frac{1}{32} = \frac{31}{32}$$

*The probability of having at most one heads equals:
$${5 \choose 0} \bigg(\frac{1}{2}\bigg)^0 \bigg(\frac{1}{2}\bigg)^5 + {5 \choose 1} \bigg(\frac{1}{2}\bigg)^1 \bigg(\frac{1}{2}\bigg)^4 = \frac{6}{32}$$
