Three cards are drawn at random from a standard deck with replacement. What is the probability that:
- Two hearts were drawn?
- three hearts were drawn?
- at least one heart was drawn?
for #1 I got $52C3 \cdot (3/13)^3 \cdot (11/13)^{52-3}$ but I don't think I am doing this correctly....now I am completely confused.
The question states that you are drawing with replacement, i.e. once you've drawn a heart, you put it back in, so the probability of drawing a heart (1 in 4) never changes. Having said that, the answers would be:
1: $\left(\frac{1}{4}\right)^2 \left(\frac{3}{4}\right) \times \ ^3 C_2$
2: $\left(\frac{1}{4}\right)^3$
3: $1 - \left(\frac{3}{4}\right)^3$
For (1): You get 2 hearts out of 3, and the third is not a heart; you could get this in $^3 C_2 = 3$ ways: the first and second are hearts, the second and third are hearts, or the first and third are hearts.
For (2): You need 3 hearts in a row.
For (3): The probability of you not not drawing a single heart (i.e. drawing only non-hearts) is $\left(\frac{3}{4}\right)^3$. The complement of this event is that you draw at least one heart, which is what you want. So subtract this probability from 1.