Unformatted text preview: 5 c) Prob(David is not dealt a K or a Q) To calculate this, count the number
of ways David is dealt a hand from the remaining cards (after removing the K's
and Q's 16 remain). This is equal to (16,5). Divide this by the total number of
ways to select 5 cards out of 24 cards.
5 d) By de Morgan's law (AUB)c = Ac Bc
A is the event that peggy is dealt no king.
B is the event that peggy is dealt no queen.
We had calculated the probability that a player was not
dealt a K or a Q in part c of question ve. Using
that and de Morgan's law, we get the answer.
5 e) Here, I made a small error and I will recalculate.
P (K
Q)
P(KQ)= P (Q)
P(K Q)= (4,1)(4,1)(22,3) The numerator is the number of ways of choosing
(24,5)
one king
and one queen from the deck and then arbitrarily choosing the remaining
cards.
P(Q)=1 − (20,5)
(24,5)
(20,5)
Probability of not getting a queen = (24,5) where the numerator is no
of ways of selecting 5 cards from a suit of 20(without queens)
so
(4,1)(4,1)(22,3)
P(KQ)= (24,,5)
(20 5)
(24,5) 5 f) This question is a little unclear because in part b we are asked to
calculate probability
of getting a queen. Whereas, in part c we are asked to calculate probability
of getting a K
given that we have a queen. If we reinterpret part b as calculating probability
of getting a K
(it makes no dierence to the calculation), we can say that answer in part e
is greater than the
answer in part b. This is because we are given information the distribution,
our probability space
has become smaller. Correspondingly, the probabilities of events increases
to keep the sum equal
to 1. 1 ...
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This note was uploaded on 01/21/2012 for the course MATH 235 taught by Professor Celmin during the Spring '08 term at Waterloo.
 Spring '08
 CELMIN

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