- #1

PhDeezNutz

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- 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

## 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

## 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

## 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?

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?