Probability Explanation And Exercises Worksheet Page 11
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37P(ace on second draw | an ace on the first draw)
The vertical bar “|” is read as “given,” so the above expression is short for: “The
probability that an ace is drawn on the second draw given that an ace was drawn on
the first draw.” What is this probability? Since after an ace is drawn on the first
draw, there are 3 aces out of 51 total cards left. This means that the probability that
one of these aces will be drawn is 3/51 = 1/17.
If Events A and B are not independent, then
P(A and B) = P(A) x P(B|A).
Applying this to the problem of two aces, the probability of drawing two aces from
a deck is 4/52 x 3/51 = 1/221.
One more example: If you draw two cards from a deck, what is the
probability that you will get the Ace of Diamonds and a black card? There are two
ways you can satisfy this condition: (a) You can get the Ace of Diamonds first and
then a black card or (b) you can get a black card first and then the Ace of
Diamonds. Let's calculate Case A. The probability that the first card is the Ace of
Diamonds is 1/52. The probability that the second card is black given that the first
card is the Ace of Diamonds is 26/51 because 26 of the remaining 51 cards are
black. The probability is therefore 1/52 x 26/51 = 1/102. Now for Case B: the
probability that the first card is black is 26/52 = 1/2. The probability that the
second card is the Ace of Diamonds given that the first card is black is 1/51. The
probability of Case B is therefore 1/2 x 1/51 = 1/102, the same as the probability of
Case A. Recall that the probability of A or B is P(A) + P(B) - P(A and B). In this
problem, P(A and B) = 0 since a card cannot be the Ace of Diamonds and be a
black card. Therefore, the probability of Case A or Case B is 1/102 + 1/102 = 2/102
= 1/51. So, 1/51 is the probability that you will get the Ace of Diamonds and a
black card when drawing two cards from a deck.
Birthday Problem
If there are 25 people in a room, what is the probability that at least two of them
share the same birthday. If your first thought is that it is 25/365 = 0.068, you will
be surprised to learn it is much higher than that. This problem requires the
application of the sections on P(A and B) and conditional probability.
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