40 playing cards. 4 aces still in the deck. If I pick two cards...?

Austix

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40 playing cards. 4 aces still in the deck. If I pick two cards what is the probability of drawing one ace exactly ?
 
40 playing cards. 4 aces still in the deck. If I pick two cards what is the probability of drawing one ace exactly ?
There are 2 ways of getting exactly one ace, ie ace first then not an ace OR not an ace first then an ace.
Can you work out the P( ace first then not an ace )?
 
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well there is a 1/10 chance of drawing an ace first, and then a 12/13 chance of drawing a non ace second-- is that right?
 
This can be done as 'ace first then non-ace' or 'non-ace first then ace'.

There are 40 cards 4 of them aces so the probability of 'ace first' is, as you say, 4/40= 1/10. There are then 39 cards left, 36 of the non-aces. The probability of drawing a non-ace second is 36/39= 12/13, also as you say. So what is the probability of 'ace first then non-ace'?

Now, go back- there are 40 cards, 36 of them non-aces. The probability of drawing a non-ace first is 36/40= 9/10. There are then 39 cards left in the deck, 4 of them aces. The probability of drawing an ace second is 4/39. So what is the probability of 'non-ace first then ace'?

(The final answer in both paragraphs should be the same.)

From that, what is the probability of 'ace first then non-ace' or 'non-ace first then ace'?
 
OK so in the first case 12/130 or in the second 36/390 and both situations equal a 9.23% of occurring. Did I get that right?
 
40 playing cards. 4 aces still in the deck. If I pick two cards what is the probability of drawing one ace exactly ?

\(\displaystyle \dfrac{\dbinom{4}{1}\dbinom{36}{1}}{\dbinom{40}{2}}=\dfrac{(4)(36)(2)(1)}{(40)(39)}\)
 
OK so in the first case 12/130 or in the second 36/390 and both situations equal a 9.23% of occurring. Did I get that right?
Yes, so now you need to add them together.

P(drawing exactly one ace) = P(ace first, then not ace) + P (not ace, then ace)

= \(\displaystyle \frac{4}{40}*\frac{36}{39} + \frac{36}{40}*\frac{4}{39}

= \frac{12}{130}+\frac{12}{130} = \frac {24}{130}= \frac{12}{65}\)
 
Yes, so now you need to add them together.
P(drawing exactly one ace) = P(ace first, then not ace) + P (not ace, then ace)
= \(\displaystyle \frac{4}{40}*\frac{36}{39} + \frac{36}{40}*\frac{4}{39}
= \frac{12}{130}+\frac{12}{130} = \frac {24}{130}= \frac{12}{65}\)
The mathematics (arithmetic) can be easier. Choose one ace, choose one non-ace.
\(\displaystyle \dfrac{\dbinom{4}{1}\dbinom{36}{1}}{\dbinom{40}{2}}=\dfrac{(4)(36)(2)(1)}{(40)(39)}=\dfrac{(12)(2)(1)}{(10)(13)}=\dfrac{12}{65}\).
 
thank you

Have decided to study mathematics late in life -- any great websites for this?
 
This can be done as 'ace first then non-ace' or 'non-ace first then ace'.

There are 40 cards 4 of them aces so the probability of 'ace first' is, as you say, 4/40= 1/10. There are then 39 cards left, 36 of the non-aces. The probability of drawing a non-ace second is 36/39= 12/13, also as you say. So what is the probability of 'ace first then non-ace'?

Now, go back- there are 40 cards, 36 of them non-aces. The probability of drawing a non-ace first is 36/40= 9/10. There are then 39 cards left in the deck, 4 of them aces. The probability of drawing an ace second is 4/39. So what is the probability of 'non-ace first then ace'?

(The final answer in both paragraphs should be the same.)

From that, what is the probability of 'ace first then non-ace' or 'non-ace first then ace'?


Thank you.
 
This can be done as 'ace first then non-ace' or 'non-ace first then ace'.

There are 40 cards 4 of them aces so the probability of 'ace first' is, as you say, 4/40= 1/10. There are then 39 cards left, 36 of the non-aces. The probability of drawing a non-ace second is 36/39= 12/13, also as you say. So what is the probability of 'ace first then non-ace'?

Now, go back- there are 40 cards, 36 of them non-aces. The probability of drawing a non-ace first is 36/40= 9/10. There are then 39 cards left in the deck, 4 of them aces. The probability of drawing an ace second is 4/39. So what is the probability of 'non-ace first then ace'?

(The final answer in both paragraphs should be the same.)

From that, what is the probability of 'ace first then non-ace' or 'non-ace first then ace'?


Thank you.
(1/10)(12/13)*2=12/65
 
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