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Here is the problem: In a game played with a standard deck of cards, each face card has a value of 10 points, each ace has a value of 1 point, and each number card has a value equal to its number. Two cards are drawn at random.

One card is the queen of diamonds. What is the probability that the sum of the cards is greater than 18?

The book says the answer is 19/51, however I got a different answer. Can someone please help me?:

HOW I GOT MY ANSWER:

Probability of (A given B) = P(Sum of two cards is greater than 18 given one is a queen of diamonds) = P(sum of 2 cards is greater than 18 and one of them is a queen of diamonds)/P(drawing a queen of diamonds)

P(drawing a queen of diamonds) = 1/52

P(sum of 2 cards is greater than 18 and one of them is a queen of diamonds) = (1/52)*(15/51) = 5/884

To get a sum greater than 18, I can draw any of the other face cards (11 other) or any card of 9 (4 of those), to get 15/51.

So to get my answer, I evaluated (5/884)/(1/52), to get 5/17, which is clearly not equal to the book answer, 19/51. If there is anything I have done wrong or missed, please tell me, thank you so much!

Also, if there is anything I need to be more clear on, don't be afraid to ask!

Cheers, Justin

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  • $\begingroup$ The other card has to be face, 10, 9 or 8. There are 7 face cards left (the QD is already taken), 4 tens, 4 nines and 4 eights. So there are 19 cards which work. There are 51 cards to choose from, so 19/51. $\endgroup$ – almagest May 7 '16 at 18:55
  • $\begingroup$ But if I chose an 8, the sum would be 18, which is not greater than 18. (And yes, I forgot about the 10 lol) $\endgroup$ – Justin May 7 '16 at 18:57
  • $\begingroup$ Oh, sorry. I though it was 18 or greater. Careless. Well in that case, the answer is 15/51=5/17 and your conclusion is correct. $\endgroup$ – almagest May 7 '16 at 18:59
  • $\begingroup$ There are 11 face card remaining? 3Qs, 4Js and 4Ks. So there are 19 favourite cards remained in the 51 cards. $\endgroup$ – BGM May 7 '16 at 19:13
  • $\begingroup$ Ok I get it now, thanks for the help guys! $\endgroup$ – Justin May 7 '16 at 19:26
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For the sum of the two values to exceed $18$ given that one of them is the queen of diamonds, the other card must have a value of $9$ or $10$ points. There are four cards that have the value $9$ points ($9\color{red}{\heartsuit}$, $9\color{red}{\diamondsuit}$, $9\clubsuit$, $9\spadesuit$). There are $16$ cards with the value $10$ points, namely, the $10$, jack, queen, and king in each of the four suits, one of which is the queen of diamonds. Thus, of the $51$ cards other than the queen of diamonds in the deck, there are $4 + 16 - 1 = 15$ cards with a value of $9$ or higher. Hence, the probability that the sum of the values is greater than $19$ given that one of them is the queen of diamonds is $19/51$.

In making your calculations, you overlooked the $10\color{red}{\heartsuit}$, $10\color{red}{\diamondsuit}$, $10\clubsuit$, $10\spadesuit$.

More formally, let $A$ be the event that a two-card hand containing the queen of diamonds is drawn. Let $B$ be the event that a two-card hand in which the sum of the values is greater than $19$ is drawn. The probability that event $B$ occurs given that event $A$ has occurred is $$P(B \mid A) = \frac{P(A \cap B)}{P(A)} = \frac{\binom{1}{1}\binom{19}{1}}{\binom{1}{1}\binom{51}{1}} = \frac{19}{51}$$ because a favorable two-card hand that contains the queen of diamonds must contain the only queen of diamonds and one of the other $19$ cards with a value of at least $9$, while a two-card hand that contains the queen of diamonds must contain the queen of diamonds and one of the other $51$ cards in the deck.

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    $\begingroup$ Great and thorough explanation, thank you! $\endgroup$ – Justin May 10 '16 at 3:03

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