What is the probability that a six card hand drawn from a standard 52 card deck (without replacement) has at least 2 hearts?
I set up cases for the chances of no hearts and 1 heart appearing, added their probabilities together, and subtracted them from 1, but I was incorrect.
Can anyone give me a "dumbed-down" explanation for this? It's a study guide question I want to understand.
Best Answer
You have the right technique. How did you calculate the probabilities?
To calculate the probability of selecting $h$ cards from the hearts, and $6-h$ cards from the other suits, out of all the ways to select $6$ cards from $52$, use:
$$\mathsf P(N_\heartsuit = h) = {\binom{13}{h}\binom{39}{6-h}\over\binom{52}{6}}$$
So: $$\begin{align}\mathsf P(N_\heartsuit \geq 2) & = 1- {\binom{39}{6}\over\binom{52}{6}}- {13\times\!\binom{39}{5}\over\binom{52}{6}} \\[5pt] & =\frac{1886}{3995} \end{align}$$