There were quite a few deck of cards probability problems and I went through a few but couldn't find anything close so please forgive me if this is a repeat.
The question is as follows:
Two cards are chosen without replacement at random from a standard 52-card deck. What is
the probability that
(b) the first card is a heart and the second card is a 10?
We will be dividing by C(52,2) since we're choosing two cards from a deck of 52. This quantity is 1326.
Our first choice must be a heart so we have C(13,1) since there are 13 hearts in a deck and we're choosing one.
Second choice must be a 10 so we need C(4,1)
Thus, we get $\frac{C(13,1)C(4,1)}{C(52,2)}$ = 3.92%
Is my logic correct?
Thanks.
Best Answer
The way that I would approach this question is to approach the two events separately and then multiply their probabilities.
So first we try to find the probability of choosing a heart out of the deck, this is pretty obvious as there are only $4$ suits, and therefore:
$$P(\text{Heart})=\frac{1}{4}$$
Next we look at the probability of choosing a $10$ considering that we've just taken a heart of unknown value out of the deck:
$$\begin{align}P(10\mid \text{Heart already taken})&=P(10\text{ of Hearts removed})P(10 \mid 10\text{ of Hearts removed}) \\ &+ P(10\text{ of Hearts not removed})P(10\mid 10\text{ of Hearts not removed})\\ &= \frac{1}{13}\times\frac{3}{51}+\frac{12}{13}\times\frac{4}{51}=\frac{1}{13}\end{align}$$
We then multiply the two probabilities together to get:
$$\begin{align}P(\text{Heart followed by a }10)&=P(\text{Heart})P(10\mid \text{Heart already taken}) \\ &= \frac{1}{4}\times\frac{1}{13}=\frac{1}{52}\approx1.92\% \end{align}$$