Conditional Probability and/or Binomial Distribution(s)

[ObliviousSage]

Homework Statement

A car dealer estimates that 50% of customers entering the dealership will buy a normal car, 20% will buy a high-end car, and 30% are just browsing. If 5 customers enter his dealership on a particular day, what is the probability that two will purchase high-end models, one will purchase a normal car, and two will buy nothing?

Homework Equations

probability A happens AND B happens = Pr(A)*Pr(B)

binomial distribution:
given n independent trials, each with probability p of success, probability of exactly k successes is (n!/k!(n-k)!)p^k(1-p)^n-k

The Attempt at a Solution

It seems too easy to just say the probability is simply .2*.2*.5*.3*.3, the product of the probabilities of each of the events. I can't do that, right?

I was thinking I need to model purchasing any car as a binomial distribution X, with n = 5 and p = 7/10, and purchasing a high end car as a binomial distribution Y with n = X, p = 2/7. Then the answer is p_X(3)*p_Y|X=3(2). Is this the right way to approach the problem?

Side Note: I have questions about a couple other problems from this assignment; is it considered tacky to have multiple questions open at the same time? If not, is it preferable for me to put them all in one thread or make a separate thread for each?

 

[Dick]

.2*.2*.5*.3*.3 is a good start. But that's the probability if you know the order the customers will enter. You have to multiply by a binomial factor that tells you how many ways five customers can be split into groups of one normal, two high end and two just browsing. I wouldn't say it's 'bad form' to group a lot of questions into one post. But it doesn't help to get responses if you overload people. Keeping the questions as small as possible would be best.

 

[ObliviousSage]

So setting it up as a pair of binomials like I described then taking p_X(3)*p_Y|X=3(2) wouldn't work?

 

[Dick]

So setting it up as a pair of binomials like I described then taking p_X(3)*p_Y|X=3(2) wouldn't work?

Out of five customers, how many ways are there to select two of one kind, one of a second kind and two of a third kind?

 

[ObliviousSage]

Out of five customers, how many ways are there to select two of one kind, one of a second kind and two of a third kind?

Um, does order matter?

If not, should be four; two (first kind) * one (second kind) * two (third kind).

If order does matter, should be 5! ways to arrange any 5 objects, divided by 2 for the missing diversity from the first kind, divided by two again for the missing diversity of the second kind, for a total of 30, right?

 

[Dick]

Um, does order matter?

If not, should be four; two (first kind) * one (second kind) * two (third kind).

If order does matter, should be 5! ways to arrange any 5 objects, divided by 2 for the missing diversity from the first kind, divided by two again for the missing diversity of the second kind, for a total of 30, right?

Order doesn't matter, does it? You just want to know how much is bought, not who buys what. 30, right! Then each of those 30 has a probability of .2*.2*.5*.3*.3, yes?

 

[ObliviousSage]

30, right! Then each of those 30 has a probability of .2*.2*.5*.3*.3, yes?

That sounds right. Hmmm, that gives me a .054 chance of it happening, which is exactly half of what I came up with when I did it with the two binomial distributions. Is there something else we need to do with our 30 different .2*.2*.5*.3*.3 chances, or did I do something wrong in the binomials?

Edit: But, if order doesn't matter, it should only be 4, right? Now I'm lost again...

 

[Dick]

That sounds right. Hmmm, that gives me a .054 chance of it happening, which is exactly half of what I came up with when I did it with the two binomial distributions. Is there something else we need to do with our 30 different .2*.2*.5*.3*.3 chances, or did I do something wrong in the binomials?

Edit: But, if order doesn't matter, it should only be 4, right? Now I'm lost again...

I think 0.054 is correct. I could be wrong, sometimes am. But I don't see why you think 4 is the number of ways of grouping the customers.

 

[ObliviousSage]

I think 0.054 is correct. I could be wrong, sometimes am. But I don't see why you think 4 is the number of ways of grouping the customers.

Never mind, I was having a brain fart. I'm still unclear on why the binomial way produces two times 0.054, though. Lemme go back and check my math on it, it may be that it's supposed to give the same answer and I just added it up wrong.

 

[ObliviousSage]

Never mind, I wasn't getting twice that 0.054, I was getting one fifth of it, 0.0108. Math still seems to check out, though. Very weird. Why wouldn't the binomial way give me the right answer?

 

[Dick]

Never mind, I wasn't getting twice that 0.054, I was getting one fifth of it, 0.0108. Math still seems to check out, though. Very weird. Why wouldn't the binomial way give me the right answer?

0.054 IS the binomial way, (number of ways to choose from the 5)*(probability of choice). It's just with three possibilities instead of two. What are you doing to get 0.0108?

 

[ObliviousSage]

As described in the first post, two binomial distributions: X represents buying any car, and has n=5, p=7/10, while Y represents someone buying a car choosing a high-end car, and has n=X and p=2/7.

Binomial distributions have probability p(k)=(n choose k)*p^k*(1-p)^n-k.

We thus want the probability of getting 3 of 5 to buy cars, and 2 of 3 car-buyers to buy high-end cars. That should be p_X(3)*p_Y|X=3(2).

Thus we get (5!/3!2!)(7/10)³(3/10)²(3!/2!1!)(2/7)²(1/7) = 12*3²/10⁴ = 0.0108.

 

[Dick]

As described in the first post, two binomial distributions: X represents buying any car, and has n=5, p=7/10, while Y represents someone buying a car choosing a high-end car, and has n=X and p=2/7.

Binomial distributions have probability p(k)=(n choose k)*p^k*(1-p)^n-k.

We thus want the probability of getting 3 of 5 to buy cars, and 2 of 3 car-buyers to buy high-end cars. That should be p_X(3)*p_Y|X=3(2).

Thus we get (5!/3!2!)(7/10)³(3/10)²(3!/2!1!)(2/7)²(1/7) = 12*3²/10⁴ = 0.0108.

The odds of buying a high-end car once you decided to buy a car is (2/7). That's ok. But then your odds of not buying a high-end car are 5/7. The last factor of 1/7 should be 5/7.

 

[ObliviousSage]

The odds of buying a high-end car once you decided to buy a car is (2/7). That's ok. But then your odds of not buying a high-end car are 5/7. The last factor of 1/7 should be 5/7.

Whoo, math derp! OK, now it comes out to 0.054, the same as the other way, which is good because they both seemed like valid ways of solving the problem. Awesome, thanks for the help!