Two cards are drawn from a deck without replacement

[PhDeezNutz]

Homework Statement

Two cards are drawn from a deck (without replacement), what is the probability that at least one card is a picture card

Relevant Equations

for non-mutually exclusive events
$P\left(X \cup Y \right) = P \left(X \right) + P \left( Y \right) - P\left( X \cap Y\right)$

for dependent events (the occurrence of one event effects the occurrence of another)

$P\left( X \cap Y\right) = P\left( X\right) P \left(Y \vert X\right)$

where $P \left(Y \vert X\right)$ means Probability of Y given X occurs

Let $X$ be the event that the first draw is a picture card
Let $Y$ by the event that the second is a picture card

Then the probability that at least one of the cards is a picture card is the probability of $X$ union $Y$ and has the formula

Equation 1
$P\left(X \cup Y \right) = P \left(X \right) + P \left( Y \right) - P\left( X \cap Y\right)$


I contend $P \left(X\right) = \frac{12}{52}$ and $P \left( Y \right) = \frac{12}{51}$ (because with one card gone there are now $51$ cards) left

however the probability of $Y$ definitely depends on $X$ and not the other way around. Given that $P\left(X \cup Y \right) = P\left(Y \cup X \right)$ Equation 1 becomes

Equation 2

$P\left(Y \cup X \right) = P \left(Y \right) + P \left( X \right) - P\left( Y \cap X\right)$

Given that $P \left ( Y \cap X\right) = P \left(X \right) P \left( Y \vert X \right)$

Equation 2 becomes
Equation 3

$P\left(Y \cup X \right) = P \left(Y \right) + P \left( X \right) - P\left( Y \cap X\right) = P \left( Y \right) + P \left( X \right) - P\left(X \right) P\left(Y \vert X \right)$


$P\left(Y \vert X \right) = \frac{11}{51}$ because if one picture card is gone that leaves $11$ picture cards out of a total of $51$ cards

Plugging in all the numbers

$\frac{12}{51} + \frac{12}{52} - \frac{12}{52} \left( \frac{11}{51}\right)$

$= \frac{93}{221}$

Apparently the correct answer is $\frac{91}{221}$

where did I go wrong?

 

[pasmith]

There are only 11 face cards out of 51 cards in the deck if the first card drawn was a face card. Thus P(Y) = \frac{12}{52}\frac{11}{51} + \frac{40}{52}\frac{12}{51} = \frac{12}{52}. You then have <br /> P(X) + P(Y) - P(X \cap Y) = \frac{12}{52} + \frac{12}{52} - \frac{11 \cdot 12}{52 \cdot 51} = \frac{1092}{2652} = \frac{7}{17} = \frac{91}{221}.

 

[Orodruin]

As the problem has been solved: A more direct computation:

Ways of drawing 2 out of 52 cards (unordered) = ${52 \choose 2} = 52\cdot 51 / 2 = 1326$
Ways of drawing 2 non-face cards (unordered) = ${40 \choose 2} = 40\cdot 39 / 2 = 780$
Probability of no face card = $780/1326 = 10/17$
Probability of at least one face card = $1 - 10/17 = 7/17 = 91/221$

 

[Orodruin]

There are only 11 face cards out of 51 cards in the deck if the first card drawn was a face card. Thus P(Y) = \frac{12}{52}\frac{11}{51} + \frac{40}{52}\frac{12}{51} = \frac{12}{52}.

Also, just to point out the obvious: There is no need to split $Y$ into the two cases conditioned on $X$. As $P(Y)$ is the probability of the second card being a face card, it should be clear that the (unconditioned) probability is 12/52 as 12 out of the 52 cards that may be second in the deck are face cards. The probability of any fixed card ending up at any particular fixed position is 1/52.

 

[pbuk]

Yet another route to the correct solution: when you see the words

what is the probability that at least one

this should trigger the thought "this is 1 - the probability that none"

The probability of no picture card on the first draw is $\frac{40}{52}$
The probability of no picture card on the second draw is $\frac{39}{51}$
The probability of no picture card on either the first or second draw is $\frac{40}{52} \frac{39}{51} = \frac{130}{221}$
The probability of at least 1 picture card in the first 2 draws is $1 - \frac{130}{221} = \frac{91}{221}$

 

[PhDeezNutz]

The over arching impression I get is that in the formula

$P\left(X \cup Y \right) = P \left(X \right) + P \left( Y \right) - P\left( X \cap Y\right)$

is that $P\left(X \right)$ and $P\left(Y \right)$ are treated as independent events
and the interdependency is treated by the "cross term" (for lack of a better term) in $P\left( X \cap Y\right)$

is this a correct interpretation?

 

[Orodruin]

First of all, $P(X)$ and $P(Y)$ are probabilities, not events. The events are $X$ and $Y$ and they are certainly not independent. However, their distributions are the same, which does not mean that they are independent. Being independent would mean, e.g, that the distribution of $Y$ would be independent of the realised value of $X$, which is not the case.