Expert: Sherry Wallin - 12/14/2011

Question
QUESTION: Hi

Drawing 3 cards from a 52 card deck, what is the probability/odd of:
a)them all being red
b) one of them being a 10.
c) is that the same as b) if the question was at least one being a 10

ta
chrissy

ANSWER: There are 26 red cards out of 52 cards, so the probability of choosing one red is 26/52 = 1/2
After you have chosen one red card there are only 25 red cards out of 51 cards so the probability of again drawing a red card is 25/51, and finally now there are only 24 red cards out of 50 cards so the probability of drawing a red card is now 24/50 = 12/25. Thus the probability of drawing 3 reds cards is 26/52 * 25/51 * 24/50.

part b is a little amibiguous because the problem does not read only one 10 or at least one 10. Certainly if two of them are 10's then at least one is a 10. I will assume it is meant that only one is a 10. So there are 4 tens in a deck of 52 cards so the probability of drawing a 10 one the first draw is 4/52 = 1/13. Then the next two draws cannot be a 10 so you remove the three 10's from the deck sort to speak and now you have 48 cards that are not 10's and 51 total cards to draw from. Finally for the third draw you have 47 cards out of 50 cards in the deck, thus the probability of drawing one ten is 4/52 * 48/51 * 47/50 =>(4*48*47)/(52*51*50). Note that order does not matter and if you were to draw the 10 on the second draw instead of the first draw you would get 48/52 * 4/51 * 47/50 = (48*4*47)/(52*51*50) which is the same probability as the first drawn 10 and will likewise be the same if the 10 was drawn on the 3rd draw.

At least one being a 10 means you could have one ten, two tens, or even three tens, so they are NOT the same.

Math Prof

---------- FOLLOW-UP ----------

QUESTION: Hi Sherry

thank you very much for your answer. very clear and easy to follow.
you presumed correctly for question b.
sorry for ambiguous wording of problem c.

my follow up questions are
Drawing 3 cards from a 52 card deck, what is the probability/odd of:
a) drawing two 10s
b) all being 10s
c) would the probability of drawing at least one 10 then be addition or product of the probability of the three events?

ta
chrissy

Answer
a)  4/52 * 3/51 * 48/50 = .0043433891 with the same reasoning as before. The first factor 4/52 is because there are four 10's in a deck of 52 cards and after drawing one 10 there are three 10's left in the deck of 51 cards thus 3/51 and finally after drawing the 2nd ten there are 50 cards left and 48 that are not tens...

b) 4/52 * 3/51 * 2/50 = 24/132600 = .0001809954751

c)  the probability of drawing at least one 10 in 3 draws, notationally P(X>=1) where X is the probability of drawing a 10 in any draw is the same as 1-P(drawing no 10's in three draws)
= 1-[(48/52)*(47/51)*(46/50)] = .217375566

Math Prof