Two playing cards are drawn from a deck, they are the 5 of Spades and the Queen of Hearts. What is the probability that the third card drawn has a value between the 5 and the Queen? (Note: suit does matter here.)
Let event $\displaystyle A=\text{drawing a card between 5 and queen on the third draw, where suit does matter}$
Let event $\displaystyle B=\text{2 cards are 5}\spadesuit\text{ and Q}\heartsuit$
So you are looking for this:
$\displaystyle P(A|B)=\frac{P(A\cap B)}{P(B)}\backepsilon P(B)\neq0$
So we see here that $\displaystyle P(A\cap B)=P(B)P(A|B)$
It should be evident that $\displaystyle P(B)=\frac{\binom{13}1\binom{13}1}{\binom{52}2}=.1 27$ and that $\displaystyle P(A|B)=\frac{30}{50}=\frac{3}{5}$, assuming that the cards are not replaced.
Thus, $\displaystyle P(A\cap B)=.127\times .6=\color{red}\boxed{.076}$
Does this make sense?
--Chris
Hello, teekayoh!
Two playing cards are drawn from a deck: $\displaystyle 5\spadesuit$ and $\displaystyle Q\heartsuit$.
What is the probability that the third card drawn has a value
between the 5 and the Queen? (Note: suit does matter here.)
. . Not sure how the suit matters.
There are 50 cards remaining in the deck.
There are 24 cards between 5 and Queen: four each of {6, 7, 8, 9, 10, J}
Therefore: .$\displaystyle P(5 < x < Q) \;=\;\frac{24}{50} \;=\;\frac{12}{25}$
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If "between" includes 5 and Q, among the 50 remaining cards,
. . there are 24 cards mentioned above, plus $\displaystyle \{5\clubsuit,\:5\heartsuit,\:5\diamondsuit,\:Q\clu bsuit,\:Q\spadesuit,\: Q\diamondsuit\} $
Therefore: .$\displaystyle P(5 \leq x \leq Q) \;=\;\frac{30}{50} \;=\;\frac{3}{5}$