MATH 1110 - Section 001
Spring 2012
Quiz 00 Name:
February 28, 2012
Directions: Write legibly, the use of documents is forbidden, giving or receiving aid on this assignement is forbidden. You will get zero if any of these happen.
1. What is the probability of getting 2 pairs. 0 points
Answer The probability of getting 2 pairs can be described by choosing 2 rank from the 13. For each rank, choose 2 cards from 4. (13)(4)(4) Thus the associated probability is: 2 522 2 . (4) 2. What is the probability of getting 1 pair then another pair. 0 points
Answer The probability of getting 1 pair then another pair can be described by choosing 1 rank from the 13. For this rank, choose 2 cards from 4. Then, choose another rank (from 12 now), and choose 2 cards from 4. (4,2)×P (4,2) (4,2)×P (4,2) Thus the associated probability is: P (13,1)×P (12,1)×P = 13×12×P . P (52,4) 52×51×50×49 Notice, in the second case the order matters since we want to differentiate the hand 10H, 10S, 8D, 8H from the hand 10H, 8D, 10S, 8H. For this reason it is important to use permutations instead of combinations. We want this first hand to be a possible outcome, whereas the second hand is not a possible outcome; this also affects the total number of outcomes (on the bottom). Note that P (13, 1) = 13 1 = 13.
MATH 1110 - Section 001
Spring 2012
Quiz 00 Name:
February 28, 2012
Directions: Write legibly, the use of documents is forbidden, giving or receiving aid on this assignement is forbidden. You will get zero if any of these happen.
1. What is the probability of getting 2 pairs. 0 points
Answer The probability of getting 2 pairs can be described by choosing 2 rank from the 13. For each rank, choose 2 cards from 4. (13)(4)(4) Thus the associated probability is: 2 522 2 . (4) 2. What is the probability of getting 1 pair then another pair. 0 points
Answer The probability of getting 1 pair then another pair can be described by choosing 1 rank from the 13. For this rank, choose 2 cards from 4. Then, choose another rank (from 12 now), and choose 2 cards from 4. (4,2)×P (4,2) (4,2)×P (4,2) Thus the associated probability is: P (13,1)×P (12,1)×P = 13×12×P . P (52,4) 52×51×50×49 Notice, in the second case the order matters since we want to differentiate the hand 10H, 10S, 8D, 8H from the hand 10H, 8D, 10S, 8H. For this reason it is important to use permutations instead of combinations. We want this first hand to be a possible outcome, whereas the second hand is not a possible outcome; this also affects the total number of outcomes (on the bottom). Note that P (13, 1) = 13 1 = 13.
