SOLUTION: Three cards are drawn at random without replacement. Without knowing the values of the first two cards, what is the probability of the third card having value of 10 and/or being a

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Question 1146498: Three cards are drawn at random without replacement. Without knowing the values of the first two cards, what is the probability of the third card having value of 10 and/or being a face card?
Found 2 solutions by solver91311, VFBundy:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


The number of face cards and 10s in your deck divided by the number of cards remaining in your deck once you have drawn two cards. The fact that either or both of the two cards already drawn might be 10s or faces doesn't matter. You don't know what they are so all of the 10s and faces are still available as far as you know.


John

My calculator said it, I believe it, that settles it

Answer by VFBundy(438) About Me  (Show Source):
You can put this solution on YOUR website!
There are 16 cards that have a value of 10: the 10, J, Q, and K of all four suits.

Chances both of the first two cards have a "10" value = 16/52 * 15/51 = 20/221

Chances first card has a "10" value and second card does not have a "10" value = 16/52 * 36/51 = 48/221

Chances first card does not have a "10" value and second card has a "10" value = 36/52 * 16/51 = 48/221

Chances neither of the first two cards have a "10" value = 36/52 * 35/51 = 105/221

Continuing...

20/221 of the time, two "10" cards have been drawn, leaving 14 "10" cards remaining out of the 50 remaining cards. Or, a 14/50 chance the third card will be a "10" card.

96/221 of the time, one "10" card has been drawn, leaving 15 "10" cards remaining out of the 50 remaining cards. Or, a 15/50 chance the third card will be a "10" card. (The 96/221 number comes from adding together the chance where a "10" card is drawn first and the chance where a "10" card is drawn second.)

105/221 of the time, zero "10" cards have been drawn, leaving all 16 "10" cards remaining out of the 50 remaining cards. Or, a 16/50 chance the third card will be a "10" card.

So...

20/221 of the time, there is a 14/50 probability the third card will be a "10" card.

96/221 of the time, there is a 15/50 probability the third card will be a "10" card.

105/221 of the time, there is a 16/50 probability the third card will be a "10" card.

The math is as follows:

(20/221) * (14/50) = 28/1105
(96/221) * (15/50) = 144/1105
(105/221) * (16/50) = 168/1105

Add together 28/1105 + 144/1105 + 168/1105 and you get 340/1105. This fraction reduces to 4/13.

So, there is a 4/13 probability the third card will be a "10" card.