Question

can someone help please

Jose is paying for a six dollar meal using bills in his wallet. He has four one dollar bills, three five dollar bills, and two ten dollar bills. If he selects two bills at random, one at a time from his wallet, what is the probability that he will choose a one dollar bill and a five dollar bill to pay for the meal? Show your work.

Answer 1

@LegendarySadist

Answer 2

can you guys help

Answer 3

I'm here to see how someone else solves it - Probability was never my best subject

Answer 4

agreed

Answer 5

i could be totally wrong but ill take a guess

Answer 6

i know this math subject sucks

Answer 7

hate probability like i do not get my teacher explains and i am like uhhhh

Answer 8

total is 4 + 3 + 2 = 9,

first pull he has 4 / 9 chance to pull one dollar bill now total bills is 8

second pull he has 3/8 chance to to pull a five dollar bill

so 4/9 * 3/8 = answer?

Answer 9

its competition who will win

Answer 10

that might be wrong i dont know

Answer 11

But what if he gets a five dollar bill then a one dollar?

Answer 12

^^

Answer 13

seems right i know you have to add first

Answer 14

How many bills does he have in total?

4 + 2 + 2 = 8

The probability he picks a $10 bill first is therefore (2/8) = (1/4)
The probability he picks a $10 bill second, given he didn't pick one first, is (2/7) because there are still two $10 bills in his wallet but only seven bills left

Similarly, the probability he picks a $1 bill first is (4/8) = (1/2)
The probability he picks a $1 bill second, given he didn't pick one first, is (4/7) because there are still four $1 bills in his wallet but only seven bills left

So the probability he picks exactly $11 out with a $1 and $10 bill is the probability either he picks $10 then $1, or $1 then $10, which is:
(1/4) * (4/7) + (1/2) * (2/7) = (1/7) + (1/7) = 2/7

Answer 15

Isn't it 4+3+2=9?

Answer 16

@nicxle.62 thats not what the question is asking

Answer 17

uhhh okay um i think @LegendarySadist is right

Answer 18

beause you do have to add

Answer 19

woah woah, I'm not saying that's the solution

Answer 20

no i know you have to add first then do the rest ya i knows but i suck at this

Answer 21

So I think I have the question figured out.

Answer 22

i know how to add this because i saw this in other questions like this but different XD

Answer 23

There are two things that can happen. Either he draws a one and then a five, or he draws a five and then a one.

Answer 24

So we have to test two things out. The chance of him getting a one then five and the chance of him getting a five then one.

Answer 25

um kind of lost :/

Answer 26

\[\frac{ 4 }{ 9 }*\frac{3}{8}\] will get you the probability of a one then five\[\frac{3}{9}*\frac{4}{8}\]will get you the probability of a five and then one.

Answer 27

so basically what i said earlier

Answer 28

ha XD

Answer 29

Yeah, somewhat. The problem I'm thinking about is whether or not we're supposed to add these two probabilities. Since the question wants whether he gets a one and five, but doesn't give an order, it could be either of these.

Answer 30

so will it be 0.16

Answer 31

on the first one

Answer 32

No need to convert to decimal form. Let's keep it in fraction form please.

Answer 33

okay so will it be 1/6

Answer 34

on both

Answer 35

Yup.

Answer 36

okay is that it or not

Answer 37

thanks for your help guys

Answer 38

@puppylife101 I did some testing to confirm my hypothesis. You ARE supposed to add the two probabilities.