Question 1052489: I tried much in this problem but I didn't get my answer correct.
The question is---
A person draws two cards successively without replacement from a pack of 52 playing cards.He tells that both cards are aces,then the probability that both cards are aces if there are 60% chances that he speaks truth-----
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the probability of drawing an ace out of the deck the first time is 4/52.
the probability of drawing an ace out of the deck the second time without replacement is 3/51.
the assumption is that you got the ace the first time and there are only 3 aces left out of 51 cards left.
so the probability of drawing two aces out of the deck without replacement is 4/52 * 3/51.
if the person speaks the truth 60% of the time, than he actually has two aces 60% of the time.
if the person lies 40% of the time, than he actually doesn't have two aces 40% of the time.
so, 60% of the time, the probability that he actually has two aces is 4/52 * 3/51, and 40% of the time, the probability that he actually has two aces is 0.
the overall probability that he actually has 2 aces is therefore .60 * 4/52 * 3/51 + .40 * 0 which is equal to .60 * 4/52 * 3/51 which is equal to .0027149321.
in fractional form, this is equal to 3/1105.
this is what i think is happening.
try it out and see if it's right.
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