QR question

[299678]

if 1/3 of the cars driven by residents in a specific town are Japanese made, what is the Probability of seeing at least one Japanese car out of every 3 automobile on the town?

I used the binomial probability rule which is nCr P^r q^n-r
so nCr = 3 x 2 x 1 / 2 x 1 = 3,
P^r = (1/3)^1 = 1/3
q^r = (2/3)^2 = 4/9

so 3 x (1/3) x 4/9 = 4/9

But correct answer is 19/27 Please correct me. Thank you

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[Streetwolf]

if 1/3 of the cars driven by residents in a specific town are Japanese made, what is the Probability of seeing at least one Japanese car out of every 3 automobile on the town?

I used the binomial probability rule which is nCr P^r q^n-r
so nCr = 3 x 2 x 1 / 2 x 1 = 3,
P^r = (1/3)^1 = 1/3
q^r = (2/3)^2 = 4/9

so 3 x (1/3) x 4/9 = 4/9

But correct answer is 19/27 Please correct me. Thank you

You're on the right track. Remember it says AT LEAST one.

Two options:

1. (Easier) Find the opposite probability, which in this case is the probability of seeing exactly 0 of them. Then 1 minus that would be the probability of seeing at least 1 car.

2. (Longer) Keep doing what you're doing but include the cases where you see 2 cars and 3 cars.

 

[299678]

Great ! street wolf thank you

 

[vr4rider]

Hi.

I don't understand why the probabilty is ~70% (or 19/27).

You have 3 cars. One of them is Japanese made.

So if you see 3 cars driving in town, doesn't that mean there's going to be a 33% chance there will be 1 Japanese car?

Please break this down for me.

 

[vr4rider]

Alrighty, so after reading it over, 33% is the chance of seeing 1 Japanese car out of 1 car.

But can someone please tell me where the 19 and 27 comes from?

Thanks.

 

[299678]

Thanks to streetwolf.
the probability not seeing zero of Japanese car is 1-1/3 = 2/3
Now we need to calculate the total probability for each 3 automobile, so
(2/3)^3 = 8/27 <- this is the probability of seeing zero out of every 3 automobiles

Now, Probability of seeing at least one Japanese car out of every 3 automobile = 1-(8/27) = 19/27

so instead of using binomial probability, this is faster way to calculate it
Hope it helps

 

[UCB05]

Hi.

I don't understand why the probabilty is ~70% (or 19/27).

You have 3 cars. One of them is Japanese made.

So if you see 3 cars driving in town, doesn't that mean there's going to be a 33% chance there will be 1 Japanese car?

Please break this down for me.

right so 33% is the chance of seeing a japanese car when you look at one car. The question wants to know what is the total chance when looking at 3 cars that you'll see 1 Japanese car out of 3, plus the chance of seeing 2 Japanese cars out of 3, plus seeing 3 Japanese cars.

 

[vr4rider]

right so 33% is the chance of seeing a japanese car when you look at one car. The question wants to know what is the total chance when looking at 3 cars that you'll see 1 Japanese car out of 3, plus the chance of seeing 2 Japanese cars out of 3, plus seeing 3 Japanese cars.

Thank you, UCB05.

That helped me understand the question and what the question is asking.

Can you please explain where the 19 and 27 comes from.

I really would like to understand this question.

Thanks~!

 

[vr4rider]

Okay, so I got this so far. J=Japanese

The chance of seeing 1J car out of 1 is .33
The chance of seeing 1J car out of 2 is .50
The chance of seeing 1J car out of 3 is 1.0

The chance of seeing 2J car out of 2 is .5/2
The chance of seeing 2J car out of 3 is .33

The chance of seeing 3J car out of 3 is 1/27

 

[Streetwolf]

J = japanese, X = other car type 1, Y = other car type 2

There needs to be 2 types of normal cars and 1 type of Japanese car.

XXX
XXY
XYX
YXX
XYY
YXY
YYX
XXJ*
XJX*
JXX*
XJJ*
JXJ*
JJX*
XYJ*
XJY*
JYX*
JXY*
YJX*
YXJ*
YYY
YYJ*
YJY*
JYY*
JJY*
JYJ*
YJJ*
JJJ*

Those are the 27 combinations you could see. Remember that X and Y are both any other car but I put two of them there to show how there's a 2/3 chance of a non-Japanese car and a 1/3 chance of a Japanese car.

Out of those, there are 19 situations where you'd see at least one Japanese car.

Mathematically:

You have 3 cars on the road that you see. You need to consider the situations where you have either 1 J car, 2 J cars, or 3 J cars.

For the 1 J car, you need to choose which car is the J car. You do this with (3 choose 1) = 3:

??J
?J?
J??

For the 2 J cars, you choose which 2 are the J cars. Use (3 choose 2) = 3:

JJ?
J?J
?JJ

For the 3 J cars, use (3 choose 3) = 1:

JJJ

Now you use the probability part. For each of the 3 possibilities with only 1 J car, the probability of the car being J is 1/3 and the probability of the car not being J is 2/3. So with 1 J car, you use (1/3) * (2/3) * (2/3) = 4/27. Multiply this by 3 because there are 3 ways the J car could appear (as shown above). You get 12/27.

For the 2 J case you use (1/3)(1/3)(2/3) = 2/27. Multiply by 3 because there are 3 ways to find 2 J cars in a group of 3 cars. You get 6/27.

For the 3 J case you use (1/3)(1/3)(1/3) = 1/27. Multiply by 1. You get 1/27.

Add them up. You get 12/27 + 6/27 + 1/27 = 19/27.

For the 1 J car case:

JXY
JYX
XJY
YJX
XYJ
YXJ
XXJ
XJX
JXX
YYJ
YJY
JYY

Those are your 12/27.

Notice how if we hadn't paid attention to where the J went, we'd have these:

JXY
JYX
JXX
JYY

Those are your 4 cases (out of 9 - didn't put the other 5 without the J).


For the 2 J case:

JJX
JJY
JXJ
JYJ
XJJ
YJJ

Those are your 6/27.

I think the 3 J case is obvious:

JJJ

Hope this helps you visualize it.

 

[vr4rider]

WOW, Streetwolf! Thank you so much!

I bet you did really good in the QR, huh?

Thanks for taking the time to write that out for me. I'm more of a visual learner so this really helped

 

[Streetwolf]

I bet you did really good in the QR, huh?

Yup. Comes easily to me.

Any other questions feel free to post. I just browse this subforum for the math questions.

 

[vr4rider]

Thank you very much. I will keep you in mind