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anonymous
 one year ago
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.
anonymous
 one year ago
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.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm here to see how someone else solves it  Probability was never my best subject

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i could be totally wrong but ill take a guess

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i know this math subject sucks

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hate probability like i do not get my teacher explains and i am like uhhhh

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0total 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
 one year ago
Best ResponseYou've already chosen the best response.0its competition who will win

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that might be wrong i dont know

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But what if he gets a five dollar bill then a one dollar?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0seems right i know you have to add first

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How 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
 one year ago
Best ResponseYou've already chosen the best response.0@nicxle.62 thats not what the question is asking

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0uhhh okay um i think @LegendarySadist is right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0beause you do have to add

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0woah woah, I'm not saying that's the solution

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no i know you have to add first then do the rest ya i knows but i suck at this

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So I think I have the question figured out.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i know how to add this because i saw this in other questions like this but different XD

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0There 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
 one year ago
Best ResponseYou've already chosen the best response.0So 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
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ 4 }{ 9 }*\frac{3}{8}\] 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
 one year ago
Best ResponseYou've already chosen the best response.0so basically what i said earlier

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, 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
 one year ago
Best ResponseYou've already chosen the best response.0No need to convert to decimal form. Let's keep it in fraction form please.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay so will it be 1/6

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay is that it or not

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thanks for your help guys

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@puppylife101 I did some testing to confirm my hypothesis. You ARE supposed to add the two probabilities.
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