# Two cards are drawn from a deck without replacement

• PhDeezNutz
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?

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 $$P(X) + P(Y) - P(X \cap Y) = \frac{12}{52} + \frac{12}{52} - \frac{11 \cdot 12}{52 \cdot 51} = \frac{1092}{2652} = \frac{7}{17} = \frac{91}{221}.$$

hutchphd, pinball1970, FactChecker and 1 other person
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##

hutchphd and PhDeezNutz
pasmith said:
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.

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

PhDeezNutz said:
what is the probability that at least one

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

The probabilty of no picture card on the first draw is ## \frac{40}{52} ##
The probabilty 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?

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.

PhDeezNutz

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