Can anyone help me solve this math problem? See below. An urn contains 2-one dollar bills, 1 five dollar bill and 1 ten dollar bill. A player draws bills one at a time without replacement from the urn until a ten-dollar bill is drawn. Then the game stops. All bills are kept by the player. Determine: (A) The probability of winning $11 (B) The probability of winning all bills in the urn. (c) The probability of the game stoping at the second draw.
(A) The probability of winning $11 How?: $1 then $10 Probability: (2/5)(1/4) = 1/10
(B) The probability of winning all bills in the urn. How?: not 10, then not 10, then not ten, then 10 Probability: (1-(1/4))(1-(1/3))(1-(1/2)1 = (3/4)(2/3)(1/2) = 1/4
(c) The probability of the game stopping at the second draw. How? anything but 10 followed by 10 Probability: (3/4)(1/3) = 1/4
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