Jose is paying for a two dollar drink using bills in his wallet. He has four one dollar bills, two five dollar bills, and two ten dollar bills. If he selects two bills at random, one at a time from his wallet, without replacement, what is the probability that he will choose two one dollar bills to pay for the drink?
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Question

Jose is paying for a two dollar drink using bills in his wallet. He has four one dollar bills, two five dollar bills, and two ten dollar bills. If he selects two bills at random, one at a time from his wallet, without replacement, what is the probability that he will choose two one dollar bills to pay for the drink?
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anonymous · Answer

whoa thats alot of money in problem

anonymous · Answer

what if i dont wanna show my work :O

anonymous · Answer

go away then

anonymous · Answer

jk

anonymous · Answer

NO you cant make me, im adorable remember? :D and i sent you a message, you never replied

anonymous · Answer

The fact that the outcome of choosing the first bill  affects the outcome of choosing the second bill makes these events dependend. ( without replacement)

E1 = Event that the first  bill is a one dollar bill.
E2 = Event that the second bill is a one dollar bill.

P(E1)  and E2)  =  P(E1) and P(E2|E1)
P(E1) and P(E2|E1) =  P(E1)*P(E2|E1)


There are 4 + 2 + 2   =  8 bills to choose from.

a) P(E1)  =  4/8  
 prob. of choosing  1 of the 4  one dollar bills from 8 bills
b)  P(E2|E1)  =  3/7
  prob. of choosing  1 of the  remaining 3 one dollar bills from remaining  7 bills

c) P(E1 and E2)   = P(E1)*P(E2|E1) =(4/8) * (3/7)  = 3/14

Answer is 3/14  = 0.214286