Illinois license plates used to consist of either three letters followed by three digits or two letters followed by four digits. Assume that all the different possible license plates are equally likely.
(a) What is the probability that a randomly chosen plate contains the number 7777? (Round your answer to six decimal places.)

(b) What is the probability that a randomly chosen plate contains the substring WE? (For example, WE4321 or PWE786 are two ways WE might appear. Round your answer to six decimal places.)

Question

Illinois license plates used to consist of either three letters followed by three digits or two letters followed by four digits. Assume that all the different possible license plates are equally likely.
(a) What is the probability that a randomly chosen plate contains the number 7777? (Round your answer to six decimal places.)


(b) What is the probability that a randomly chosen plate contains the substring WE? (For example, WE4321 or PWE786 are two ways WE might appear. Round your answer to six decimal places.)

kc_kennylau · Answer

|dw:1384928810330:dw|

kc_kennylau · Answer

For (a), we only look at the last 4 places.
For (b), we only look at the first 3 places.

anonymous · Answer

So a.) is 4^10 correct?

kc_kennylau · Answer

probability is less than 1

anonymous · Answer

So what would the equation be for a.) ?

kc_kennylau · Answer

What is the probability that the 3rd place is 7?

anonymous · Answer

I'm asking you. I'm confused on this

kc_kennylau · Answer

How many cases are there for the 3rd place

anonymous · Answer

3^10

kc_kennylau · Answer

no just the 3rd place

anonymous · Answer

1

kc_kennylau · Answer

and how many possibilities are there for the 3rd place (it can be A or B or C or 1 or 3 or 7)

anonymous · Answer

7

anonymous · Answer

Your really confusing me right now

kc_kennylau · Answer

ahh nvm so the probability that the 3rd term is 7 is $\dfrac1{36}$ since there are 26 letters in the alphabet and 10 digits.

sorry for confusing you:/

anonymous · Answer

Okay now I'm getting it

kc_kennylau · Answer

The probability that the 3rd term is 7 is $\dfrac1{36}$.
The probability that the 4th term is 7 is $\dfrac1{10}$.
The probability that the 5th term is 7 is $\dfrac1{10}$.
The probability that the 6th term is 7 is $\dfrac1{10}$.

kc_kennylau · Answer

And combine it you get?

kc_kennylau · Answer

them*

anonymous · Answer

1/36 + 1/10 + 1/10 + 1/10 now I see, thank you so much

kc_kennylau · Answer

sorry don't add them together

kc_kennylau · Answer

It's the probability that the 3rd place AND the 4th AND the 5th AND the 6th is 7.
not the probability that ... OR ... OR ... OR ... is 7.

anonymous · Answer

I must multiply, yes I figured

kc_kennylau · Answer

And could you do $(b)$ by yourself? :)

anonymous · Answer

I think I understand b from your explanation of a.)

kc_kennylau · Answer

good :)
try to do it here :)

anonymous · Answer

BTW a.) keeps giving me an incorrect score

anonymous · Answer

Actually forget it, time limit on my homework ran out, thanks for all the help anyway dude

kc_kennylau · Answer

sorry :/

anonymous · Answer

Its np, I appreciate all the tutoring you did

kc_kennylau · Answer

thx :)