# Thread: Statistics Card Problem

1. ## Statistics Card Problem

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.)

2. Originally Posted by teekayoh
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 $A=\text{drawing a card between 5 and queen on the third draw, where suit does matter}$
Let event $B=\text{2 cards are 5}\spadesuit\text{ and Q}\heartsuit$

So you are looking for this:

$P(A|B)=\frac{P(A\cap B)}{P(B)}\backepsilon P(B)\neq0$

So we see here that $P(A\cap B)=P(B)P(A|B)$

It should be evident that $P(B)=\frac{\binom{13}1\binom{13}1}{\binom{52}2}=.1 27$ and that $P(A|B)=\frac{30}{50}=\frac{3}{5}$, assuming that the cards are not replaced.

Thus, $P(A\cap B)=.127\times .6=\color{red}\boxed{.076}$

Does this make sense?

--Chris

3. ## ~

Hello, teekayoh!

Two playing cards are drawn from a deck: $5\spadesuit$ and $Q\heartsuit$.
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: . $P(5 < x < Q) \;=\;\frac{24}{50} \;=\;\frac{12}{25}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

If "between" includes 5 and Q, among the 50 remaining cards,
. . there are 24 cards mentioned above, plus $\{5\clubsuit,\:5\heartsuit,\:5\diamondsuit,\:Q\clu bsuit,\:Q\spadesuit,\: Q\diamondsuit\}$

Therefore: . $P(5 \leq x \leq Q) \;=\;\frac{30}{50} \;=\;\frac{3}{5}$