Sometimes GRE will throw these AT LEAST / AT MOST Questions dealing with combinatorics and probability.

Example

John flips a fair coin four times. What is the probability that he receives at least one heads?

Solving The Looooong Way

We could calculate the probability of every case that makes us "happy" and then add those probabilities to get the answer.

Case 1: HTTT
- $\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{4!}{1!3!} = \frac{4}{16}$
Case 2: HHTT
- $\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{4!}{2!2!} = \frac{6}{16}$
Case 3: HHHT
- $\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{4!}{3!1!} = \frac{4}{16}$
Case 1: HHHH
- $\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times = \frac{1}{16}$

Add those four cases up

$\frac{4}{16} + \frac{6}{16} + \frac{4}{16} + \frac{1}{16} =$ $\frac{15}{16}$

Solving the SHORT Way

Let's simply calculate the case or cases that make us NOT "happy" and subtract it or them from $1$.

the ONLY case that makes us "unhappy": TTTT
- $\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{16}$

Subtract that from $1$

$1 - \frac{1}{16} =$ $\frac{15}{16}$

Another Example

The probability of rain on any given day is $0.25$. What is the probability that it rains AT MOST four days in a $5$-day period?

Step 1: Find the probability of the case or cases that make us unhappy.
- Only one case makes us unhappy: RRRRR

$$(0.25)^5 = \left(\frac{1}{4}\right)^5 = \frac{1}{1024}$$

$1 - \frac{1}{1024} =$ $\frac{1023}{1024}$