Again, go through the same processes
Let
S=event that the first card drawn is a spade
~S=event that the first card drawn is not a spade
s=event that the second card drawn is a spade
~s=event that the second card drawn is not a spade
then
probability of NOT drawing a spade in the second draw, given the first card drawn was a spade
P(~s|S)=39/51
probability of not drawing a spade in the second draw, given the first card drawn was NOT a spade
P(~s|~S)=38/51
We already know that the probability of drawing a space in the first card is
P(S)=13/52=1/4
Total probability of NOT drawing a spade in the second draw (any of diamond, club, heart)
P(~s)=P(~s|S)P(S)+P(~s|~S)P(~S)
=39/51*(1/4)+38/51*(3/4)
=3/4
Using Bayes theorem
P(S|~s)=P(~s|S)*P(S)/P(~s)
=(39/51)*(1/4)/(3/4)
=13/51
(slightly better than the intuitive value of 13/52=1/4).
WHY WOULD THE PROBABILITY OF THE FIRST CARD DRAWN BE AFFECTED BY THE POSTERIOR INFORMATION THAT THE SECOND CARD IS NOT A SPADE?
Imagine the extreme case where we know the following 39 draws are NOT spades, then we know that the first card drawn MUST be a spade, with probability 1.
Therefore posterior information can affect the probability of the first card drawn.
Two cards are drawn in succession without replacement from a standard deck of 52 cards. What is the probability that the first card is a spade given that the second card is a diamond? (Round your answer to three decimal places.)
1 answer
1 answer