# Thread: Why do I get a different answer using the binomial distribution function?

1. ## Why do I get a different answer using the binomial distribution function?

The question I am currently working with is as follows:

If three cards are randomly selected from a standard 52 card deck of playing cards, what is the probability that at least one of them will be a painted face card (king, queen or jack)?

Initially, I just went ahead and solved it as follows:

P(at least 1 will be a face card) = 1 - (40/52)*(39/51)*(38/50) = 55.29%

However, I went to solve it using the binomial distribution function just out of curiosity and got a slightly different answer:

P(at least 1 will be a face card) = 1 - (3 choose 3) * (40/52)^3 * (12/52)^0 = 54.48%

Why?

2. ## Re: Why do I get a different answer using the binomial distribution function?

Originally Posted by terrorsquid
The question I am currently working with is as follows: If three cards are randomly selected from a standard 52 card deck of playing cards, what is the probability that at least one of them will be a painted face card (king, queen or jack)?
Initially, I just went ahead and solved it as follows:

P(at least 1 will be a face card) = 1 - (40/52)*(39/51)*(38/50) = 55.29%

However, I went to solve it using the binomial distribution function just out of curiosity and got a slightly different answer: P(at least 1 will be a face card) = 1 - (3 choose 3) * (40/52)^3 * (12/52)^0 = 54.48%
The binomial distribution cannot be use here. The cards are not independent events.

3. ## Re: Why do I get a different answer using the binomial distribution function?

Hello, terrorsquid!

If three cards are randomly selected from a standard 52 card deck of cards,
what is the probability that at least one will be a face card (K, Q or J)?

Initially, I just went ahead and solved it as follows:

P(at least 1 face card) = 1 - (40/52)*(39/51)*(38/50) = 55.29%

However, I solved it using the binomial distribution function just out of curiosity
and got a slightly different answer:

P(at least 1 face card) = 1 - (3 choose 3) * (40/52)^3 * (12/52)^0 = 54.48%

Why?

Plato is correct . . . The draws are not independent.

With the binomial function, you are using $p = \tfrac{40}{52}$ .for each draw.
. . You are drawing the 3 cards with replacement.

Here's another approach . . .

There are 12 Face cards and 40 Others.

Select 3 cards from the 52 cards.
. . There are: . ${52\choose3} \:=\:22,\!100\text{ ways.}$

Select 3 cards from the 40 Others.
. . There are: . ${40\choose3} \:=\:9,\!880\text{ ways.}$

$\text{Hence: }\:P(\text{no Face cards}) \;=\;\frac{9,\!880}{22,\!110} \:=\:\frac{38}{85}$

$\text{Therefore: }\:P(\text{at least one Face card}) \;=\;1 - \frac{38}{85} \;=\;\frac{47}{85}$