anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
@LegendarySadist
anonymous
  • anonymous
can you guys help
anonymous
  • anonymous
I'm here to see how someone else solves it - Probability was never my best subject

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More answers

anonymous
  • anonymous
agreed
anonymous
  • anonymous
i could be totally wrong but ill take a guess
anonymous
  • anonymous
i know this math subject sucks
anonymous
  • anonymous
hate probability like i do not get my teacher explains and i am like uhhhh
anonymous
  • anonymous
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?
anonymous
  • anonymous
its competition who will win
anonymous
  • anonymous
that might be wrong i dont know
anonymous
  • anonymous
But what if he gets a five dollar bill then a one dollar?
anonymous
  • anonymous
^^
anonymous
  • anonymous
seems right i know you have to add first
anonymous
  • anonymous
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
anonymous
  • anonymous
Isn't it 4+3+2=9?
anonymous
  • anonymous
@nicxle.62 thats not what the question is asking
anonymous
  • anonymous
uhhh okay um i think @LegendarySadist is right
anonymous
  • anonymous
beause you do have to add
anonymous
  • anonymous
woah woah, I'm not saying that's the solution
anonymous
  • anonymous
no i know you have to add first then do the rest ya i knows but i suck at this
anonymous
  • anonymous
So I think I have the question figured out.
anonymous
  • anonymous
i know how to add this because i saw this in other questions like this but different XD
anonymous
  • anonymous
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.
anonymous
  • anonymous
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.
anonymous
  • anonymous
um kind of lost :/
anonymous
  • anonymous
\[\frac{ 4 }{ 9 }*\frac{3}{8}\] 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.
anonymous
  • anonymous
so basically what i said earlier
anonymous
  • anonymous
ha XD
anonymous
  • anonymous
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.
anonymous
  • anonymous
so will it be 0.16
anonymous
  • anonymous
on the first one
anonymous
  • anonymous
No need to convert to decimal form. Let's keep it in fraction form please.
anonymous
  • anonymous
okay so will it be 1/6
anonymous
  • anonymous
on both
anonymous
  • anonymous
Yup.
anonymous
  • anonymous
okay is that it or not
anonymous
  • anonymous
thanks for your help guys
anonymous
  • anonymous
@puppylife101 I did some testing to confirm my hypothesis. You ARE supposed to add the two probabilities.

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