Quiz 3 Question 6
LauraHYu
The wording is a little confusing. What do you mean by, "exactly two of the cards are kings given that at least one is a king"? I would have thought that since two of the three cards drawn must be kings, then the statement "a least one is a king" has already been fulfilled. Would you please clarify?
weisbart: Feb. 6, 2015, 8:47 p.m.
Normally, you would ask the question: What is the probability that exactly two of the cards are kings. However, you happen to know that at least one of the cards is a king. In this case, what is the probability that at least two are kings? Knowledge that at least one is a king should change your odds that exactly two are kings since you have restricted your sample space. The case of no kings is no longer a possibility. Does this help?
AZobi: Feb. 7, 2015, 8:07 p.m.
I need help on this one. I got that the probability of the first card being a king is (4/52) because there are 4 kings in the deck. The probability that the second card is a king will then be (3/51). This is because one card has already been taken out--resulting in the 51 as can be seen in the denominator. The 3 comes from the assumption that the first card is a king. The third probability is then (46/50). I got 46 after assuming that the first two cards are kings. Furthermore, the third card cannot be a king because of the word "exactly." Ergo, there are only 46 options left out of 50. Lastly, I divided by 3 factorial in order to de-order.
AZobi: Feb. 7, 2015, 8:11 p.m.
Actually, now that i look at it, I think I multiplied by 3!. In other words, my answer was [(4 choose 1)(3 choose1) (46)]/ (52 choose 3)
Alan_Mendoza: Feb. 7, 2015, 9 p.m.
I used the formula for conditional probability in an equally likely sample space. So the numerator would be the # of hands with exactly 2 kings intersect having at least 1 king, which essentially is just # of hands with exactly 2 kings. I found that to be 13 choose 1 (value of card, king) times 4 choose 2 (suit for the pair of kings) times 48 choose 1 (last card). I then divided that by the # of ways to have at least 1 king in a 3 card hand. I split that into 3 disjoint partitions: a hand with 1 king, a hand with 2 kings, and a hand with 3 kings. Therefore, it would give (13 choose 1 x 4 choose 1 x 48 choose 2) + (13 choose 1 x 4 choose 2 x 48 choose 1) + (13 choose 1 x 4 choose 3). Please correct any mistakes.
dana: Feb. 7, 2015, 9:07 p.m.
I agree with Alan!
timothyho: Feb. 8, 2015, 1:33 a.m.
Regarding Alan's question, wouldn't 13 choose 1 for the numerator make you overcount? Since now your calculating also the different ways 12 other additional values can have 2 suits of the same value with a random 3rd card.
timothyho: Feb. 8, 2015, 1:34 a.m.
you're*
weisbart: Feb. 8, 2015, 10:16 a.m.
Alan_Mendoza, I think you almost have it. However, where is the $13\choose 1$ coming from. Don't you know already that the value is a king? There is only a choice of suit or suits. \[\] Here is another possibility: Instead of using all cases, could you find the number of ways that no kings are chosen and then subtract that from the total number of possibilities? Do you get the same answer from both approaches?
Alan_Mendoza: Feb. 8, 2015, noon
I worked it out without choosing the value and I got the same answer, so I guess I did do an unnecessary step.
weisbart: Feb. 8, 2015, 2:56 p.m.
So, is there now an agreed upon answer?
AZobi: Feb. 8, 2015, 3:15 p.m.
So is the answer (4 choose 2)(48 choose 1) / [(4 choose 1)(48 choose 2) +(4 choose 2)(48 choose 1) +(4 choose 3)(48 choose 0)]?
Soniakumr: Feb. 8, 2015, 3:44 p.m.
Why is it not (3 choose 2)(4 choose 2) (48 choose 1)/ (51 choose 2). (3 choose 2) accounting for choosing 2 out of 3 cards to be kings. (4 choose 2) to pick two suits of the kings. (48 choose 1) to account for the third card. (51 choose 2) to account for the totally sample space since one card is automatically a king.
AZobi: Feb. 9, 2015, 11:20 a.m.
For my answer. The (4 choose 2) came from picking the 2 kings. The (48 choose 1) is the number of ways you can pick one other card that is not a king out of the remaining 48. Because there are a total of 52 cards, removing the kings (4 cards) will give you 48. The denominator accounts for picking one king and two non-king cards, two kings and one non-king card, and 3 kings and no non-king cards. I used the conditional probability formula for this because it is given that at least one is a king.