Question

Of the new cars in a car dealer's lot, 1 in 6 are white. Today, 4 cars were sold.

A) What is the probability that 3 of the cars sold were white?
B) What is the probability that at least 2 of the cars sold were white?

Answer 1

@tcarroll010 Do you know anything about probability? lol

Answer 2

Are you familiar with the binomial probability theorem?

Answer 3

Not really

Answer 4

so you haven't learned it yet in class?

Answer 5

No, he hasn't taught us that.

Answer 6

Wait, I think I know what you're talking about.

Answer 7

oh you have heard about it?

Answer 8

I think I've seen it before, but we have never used it in class.

Answer 9

so you've seen in your book then?

Answer 10

Yeah

Answer 11

So you've seen the following formula before

\[\Large P(X = k) = _{n}C_{k}p^k(1-p)^{n-k}\]

Answer 12

I've seen it, but I'm not quite sure how it works.

Answer 13

In this case, p = 1/6 since the probability of selecting a white care on each trial is 1/6

Answer 14

n = 4 because there are 4 cars sold (ie 4 trials done)

Answer 15

and for part a) k = 3 because we want to know the probability of selling exactly k = 3 white cars

Answer 16

So

\[\Large P(X = k) = _{n}C_{k}p^k(1-p)^{n-k}\]

becomes

\[\Large P(X = 3) = _{4}C_{3}\left(\frac{1}{6}\right)^{3}\left(1-\frac{1}{6}\right)^{4-3}\]

Answer 17

I'm writing this down as we go. I got all of that so far. What do you do next?

Answer 18

Then, you use your calculator (or you can compute by hand, let me know if you want to see it by hand) to compute 4 C 3

4 C 3 = 4

Then you use a calculator to compute

4*(1/6)^(3) * (1 - 1/6)^(4-3)

to get

0.01543209876543


So \[\Large P(X = 3) \approx 0.01543209876543\]

Answer 19

Side note: Jim I know you know what I'm about to say, so this if for Haley's benefit. In the use of the binomial theorem, independence is assumed, meaning that each event is independent of the outcome of the other events. Aside from the math, so you can understand the application of this formula, there is a big assumption going on and that is that the dealer has a HUGE number of cars so that 1/6 and 5/6 are approximated as close to that for each trial repetition. Strictly speaking, the binomial theorem can be used here with the understanding that it becomes an approximation only because of the huge number of cars. If the dealer had, say 40 cars, the binomial would be too far off. If the dealer has like 500 cars, the binomial gives a good approximation. This is a problem in "without replacement", so caution is to be taken with results in that approx is assumed. I'm sure your teacher wants to use this here because we are not given the # of cars, just assuming a LOT.

Answer 20

I think I got it now! Thanks! He never taught us this before, I'm not sure why though.

Answer 21

Do you use the same formula for the second problem?

Answer 22

yes, but you need to compute

P(X = 2)
P(X = 3)
P(X = 4)

then add up the 3 individual probabilities

Answer 23

n and p will remain the same

Answer 24

Is k=2?

Answer 25

for P(X = 2), k = 2, yes

Answer 26

for P(X = 3), k = 3,

etc etc

Answer 27

I tried it and I messed up somewhere. I'm still kind of confused with the formula.

Answer 28

what did you get for P(X = 2)

Answer 29

I'm pretty sure its wrong, but I got .0231481481

Answer 30

you should have gotten 0.1157407

Answer 31

I'm going to try again.
N= 4 P=1/6 K=2, right?

Answer 32

exactly

Answer 33

I got 0.1157407407 for P(X=2) now. I'm going to redo P(X=4)

Answer 34

ok great

Answer 35

So far I have P(X=4)=1*1/6^4*(1-1/6)^(4-4)
Is that right?

Answer 36

so far, so good

Answer 37

For 1/6^4 I got this....

Answer 38

that can be written as 0.0007716

Answer 39

either are correct

Answer 40

now add up what you got for P(X = 2), P(X = 3), and P(X = 4)

Answer 41

I got .13

Answer 42

or, more accurately, 0.131944444444444

but if you round to the nearest hundredth, then you'll get 0.13

Answer 43

Thank you!

Answer 44

yw