Question

12% of all drivers do not have a valid drivers license, 6% of all drivers have no insurance, and 4% have neither

what is the probability that a randomly selected driver either fails to have a valid license or fails to have insurance is about
a) 0.18 b) 0.2 c) 0.22 d) 0.072 e) 0.14

Answer 1

you are looking for
\[P(A\cup B)=P(A)+P(B)-P(A\cap B)\]
A event that a driver fails to have a valid licence
B event that a driver fails to have insurance
so \[P(A)=\frac{12}{100}=0.12\\ P(B)=\frac{6}{100}=0.06\\ P(A\cap B)=0.12\times 0.06=0.0072\]
do the rest...

Answer 2

perhaps i assumed not independent ?!

Answer 3

\(P(A\cap B)\) is given to be 0.04.

Answer 4

yea i just realized

Answer 5

i get 0.176 as the prob

Answer 6

did they round to 0.18

Answer 7

or did i assume wrong situation

Answer 8

It's 0.04, and not 0.004, you were almost there.

Answer 9

oh then my bad heheh
i don't know where did i add one zero lol

Answer 10

got 0.14

Answer 11

made me think that i assumed wrong situation lol

Answer 12

Your original formula is still correct.
I usually let the OP to work out the numbers, otherwise she doesn't get anything out of the exercise.

Answer 13

thanks for the correction
yeah my intention was to let her work it out
but then got this error lol

Answer 14

No problem, I know of the complications! :)

Answer 15

:)

Answer 16

so @trice2964 did you get this

Answer 17

yes! thank you! so basically you add up all the percentages, 12%+6%+4% which equals 0.22 and then you divide 0.04/.22 and the answer is about .18? @xapproachesinfinity

Answer 18

oh no
i did 12% + 6% - 4%
which give me 0.14 not 0.18

Answer 19

see the first formula that i wrote

Answer 20

yes but why do you suntract 4% instead of adding it? @xapproachesinfinity

Answer 21

because if we have two events
and we want to know the probability of either one occurring
we add the probability of each event
but if we did that we over counted so we subtract the probability of both happing at the same time
this for the independent events
\[P(A\cup B)=P(A)+P(B)-B(A \cap B)\]
that exactly what this formula says

Answer 22

and we said that \[P(A\cap B)=0.04\]

Answer 23

i meant happening not happing lol

Answer 24

perhaps van diagram can help visualize this hehe

Answer 25

exactly :)

Answer 26

wait where did the 2% and 8% percent come from?! I'm sorry you guys don't think I'm a idiot lol. I've been struggling with probability really badly, I think i think to hard about it because what I originally did was just add 0.12+0.06+0.04=.22 and then I did 0.04/.22 which equals 0.18. But I see where I was wrong now kind of lol

Answer 27

so we have 12% chance for a driver with n valid licence
also 6% chance that the driver has no insurance
lastly, the chance a drives does not have both insurance and a valid license is 4%
12%-4%=8%
6%-4%=2%
4% is the a shared percentage

Answer 28

the more you struggle and the efforts you out the more you understand :)
keep it up

Answer 29

Ohhhh I understand it a little bit better now, you subtract the 4% because its shared? @xapproachesinfinity

Answer 30

yes exactly

Answer 31

if we didn't do we over counted then

Answer 32

and if it's not shared you would add it? @xapproachesinfinity

Answer 33

if there is no shared parts
then you add straight depends on the problem

Answer 34

that happen in mutually exclusive events

Answer 35

see here
http://www.mathsisfun.com/data/probability-events-mutually-exclusive.html

Answer 36

okay! well let me try this and tell me if I it's correct, but not the right answer because I want to try and get it myself lol. But Its the same scenario
-The probability that a randomly selected driver fails to have insurance, given that he fails to have a valid license is about
a)0.67 b)0.33 c) 0.06 d)0.50 e).01

Would if be 0.06? because I subtracted the 12% from the 6%
@xapproachesinfinity

Answer 37

oh no this one is different

Answer 38

this conditional probability

Answer 39

this has the formula \[P(A/B)=\frac{P(A\cap B)}{P(B)}\]

Answer 40

conditional probabilities are different lol

Answer 41

let A be event that a driver fails to have insurance
B be event that a driver fails to have a valid license

Answer 42

okay I'm going to work the problem out now

Answer 43

okay!

Answer 44

@xapproachesinfinity okay i think i have it now! lol
So since it is conditional probability which means it can be affected by previous effects! Using the formula I divided 0.06 by 0.12 and I got d 0.50!

Answer 45

should be 0.04/0.12

Answer 46

oh my gosh . i give up =((((

Answer 47

remember that shared part that is what goes into numerator

Answer 48

oh wait you use 0.04 because the person doesnt have drivers licenses or insurance!

Answer 49

oh trying i better than nothing
and none of got it just like that heheh
constant practice....

Answer 50

no P(a and b) means a driver doesn't have good license and insurance both at one time

Answer 51

watch some videos perhaps that would help you
youtube has some good videos

Answer 52

I still get stuck on probability heheh so don't get frustrated lol

Answer 53

Maybe so! I've never been good at word problems though lol. But thank you for helping me!!!

Answer 54

no problem
probability is a fun subject though lol
just practice, we all start at nothing but we continue to do more and more problems
and that comes to be helpful :)

Answer 55

to work a word problem you need a strategy
1) read effectively the problem (
2) understand what is asked
3) think about what you know
4) make connection
5) start attacking the problem
6) solve and review

Answer 56

i'm sure you can find a good strategy that is much better :)

Answer 57

organizing the ideas is key to problem solving

Answer 58

I get confused because I don't know what formula to use for each problem and I get confused on what to put into the formula!

Answer 59

yeah i understand you :)
review the notes of the lecture before doing problems of course
see a lot of example of how to use the formulas after that do problems

Answer 60

okay for the last problem! why would you divide 0.06 by 0.12 because in the problem it says the probability of a driver who fails to have insurance, and a driver who fails to have a drivers licenses, so those percentages would be 6% and 12%

Answer 61

@xapproachesinfinity

Answer 62

first its not 0.06/0.12 it is 0.04/0.12
and that's because your question said a driver fails to have insurance , "given" that a driver fails to have valid license

Answer 63

highlight "given"

Answer 64

we are looking for the probability that a driver doesn't have insurance given that we know he does not have a valid license

Answer 65

and that's we define as \[P(A/B)=\frac{P(A\cap B)}{P(B)}\]

Answer 66

we know that \[P(A\cap B)=0.04\]

Answer 67

"Given" means it has already happened.
"So given he fails to have valid licence" means you choose those that do not have valid insurance (4%) (event A \( \cap \) B) \(out~ of\) the 12% (event B)