There are 2 ways of getting exactly one ace, ie ace first then not an ace OR not an ace first then an ace.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 ?
Yes, so now you need to add them together.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?
The mathematics (arithmetic) can be easier. Choose one ace, choose one non-ace.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)
= \(\displaystyle \frac{4}{40}*\frac{36}{39} + \frac{36}{40}*\frac{4}{39}
= \frac{12}{130}+\frac{12}{130} = \frac {24}{130}= \frac{12}{65}\)
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=12/65This 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.