## 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.

1. anonymous

2. anonymous

can you guys help

3. anonymous

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

4. anonymous

agreed

5. anonymous

i could be totally wrong but ill take a guess

6. anonymous

i know this math subject sucks

7. anonymous

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

8. 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?

9. anonymous

its competition who will win

10. anonymous

that might be wrong i dont know

11. anonymous

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

12. anonymous

^^

13. anonymous

seems right i know you have to add first

14. 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

15. anonymous

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

16. anonymous

@nicxle.62 thats not what the question is asking

17. anonymous

uhhh okay um i think @LegendarySadist is right

18. anonymous

beause you do have to add

19. anonymous

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

20. anonymous

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

21. anonymous

So I think I have the question figured out.

22. anonymous

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

23. 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.

24. 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.

25. anonymous

um kind of lost :/

26. 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.

27. anonymous

so basically what i said earlier

28. anonymous

ha XD

29. 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.

30. anonymous

so will it be 0.16

31. anonymous

on the first one

32. anonymous

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

33. anonymous

okay so will it be 1/6

34. anonymous

on both

35. anonymous

Yup.

36. anonymous

okay is that it or not

37. anonymous

38. anonymous

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

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