Three cards are randomly chosen without replacement from an ordinary deck of 52 playing cards.

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Three cards are randomly chosen without replacement from an ordinary deck of 52 playing cards.
Given that the ace of spades is chosen, what is the probability
that all three cards are aces?



Using conditional probability
$P(A|B) = P(AB)/P(B)$



$P(A) =$ cards are aces



$P(B) =$ ace of spade chosen



I ended up at $P(AB) = dfrac^3C_1^52C_3$

$3$ ways = Spade,club,heart, spade,heart,diamond and spade club diamond.



$P(B) = (^51C_2)/(^52C_3)$ assume ace of spade chosen, chose another 2 cards



$P(A|B) = [^3C_1/^52C_3)]/[^51C_2)/(^52C3_)]$



I have looked around for this question but could only find with 2 cards not 3.
Not sure if this is the correct way ?




probability proof-verification




share|cite|improve this question

















  • 2




    I think that looks fine as it is.
    – Allen O'Hara
    Jul 17 at 19:09














up vote
0
down vote

favorite












Three cards are randomly chosen without replacement from an ordinary deck of 52 playing cards.
Given that the ace of spades is chosen, what is the probability
that all three cards are aces?



Using conditional probability
$P(A|B) = P(AB)/P(B)$



$P(A) =$ cards are aces



$P(B) =$ ace of spade chosen



I ended up at $P(AB) = dfrac^3C_1^52C_3$

$3$ ways = Spade,club,heart, spade,heart,diamond and spade club diamond.



$P(B) = (^51C_2)/(^52C_3)$ assume ace of spade chosen, chose another 2 cards



$P(A|B) = [^3C_1/^52C_3)]/[^51C_2)/(^52C3_)]$



I have looked around for this question but could only find with 2 cards not 3.
Not sure if this is the correct way ?




probability proof-verification




share|cite|improve this question

















  • 2




    I think that looks fine as it is.
    – Allen O'Hara
    Jul 17 at 19:09












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Three cards are randomly chosen without replacement from an ordinary deck of 52 playing cards.
Given that the ace of spades is chosen, what is the probability
that all three cards are aces?



Using conditional probability
$P(A|B) = P(AB)/P(B)$



$P(A) =$ cards are aces



$P(B) =$ ace of spade chosen



I ended up at $P(AB) = dfrac^3C_1^52C_3$

$3$ ways = Spade,club,heart, spade,heart,diamond and spade club diamond.



$P(B) = (^51C_2)/(^52C_3)$ assume ace of spade chosen, chose another 2 cards



$P(A|B) = [^3C_1/^52C_3)]/[^51C_2)/(^52C3_)]$



I have looked around for this question but could only find with 2 cards not 3.
Not sure if this is the correct way ?




probability proof-verification




share|cite|improve this question













Three cards are randomly chosen without replacement from an ordinary deck of 52 playing cards.
Given that the ace of spades is chosen, what is the probability
that all three cards are aces?



Using conditional probability
$P(A|B) = P(AB)/P(B)$



$P(A) =$ cards are aces



$P(B) =$ ace of spade chosen



I ended up at $P(AB) = dfrac^3C_1^52C_3$

$3$ ways = Spade,club,heart, spade,heart,diamond and spade club diamond.



$P(B) = (^51C_2)/(^52C_3)$ assume ace of spade chosen, chose another 2 cards



$P(A|B) = [^3C_1/^52C_3)]/[^51C_2)/(^52C3_)]$



I have looked around for this question but could only find with 2 cards not 3.
Not sure if this is the correct way ?





probability proof-verification





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 17 at 20:15









bjcolby15

8171716




8171716









asked Jul 17 at 18:49









Sacha E

376




376







  • 2




    I think that looks fine as it is.
    – Allen O'Hara
    Jul 17 at 19:09












  • 2




    I think that looks fine as it is.
    – Allen O'Hara
    Jul 17 at 19:09







2




2




I think that looks fine as it is.
– Allen O'Hara
Jul 17 at 19:09




I think that looks fine as it is.
– Allen O'Hara
Jul 17 at 19:09










1 Answer
1






active

oldest

votes

















up vote
3
down vote



accepted










Since you have chosen ace of spades, in fact three more aces are remaining of which two more aces should be chosen the the probability is $$P=dfracbinom32binom512$$no complexity or conditional probability is needed at all.






share|cite|improve this answer





















  • Thanks, yeah i was forced to use conditional in this problem, but your answer helped me check my answer is correct ! thanks
    – Sacha E
    Jul 17 at 19:36






  • 1




    You're welcome!!
    – Mostafa Ayaz
    Jul 17 at 19:36










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
3
down vote



accepted










Since you have chosen ace of spades, in fact three more aces are remaining of which two more aces should be chosen the the probability is $$P=dfracbinom32binom512$$no complexity or conditional probability is needed at all.






share|cite|improve this answer





















  • Thanks, yeah i was forced to use conditional in this problem, but your answer helped me check my answer is correct ! thanks
    – Sacha E
    Jul 17 at 19:36






  • 1




    You're welcome!!
    – Mostafa Ayaz
    Jul 17 at 19:36














up vote
3
down vote



accepted










Since you have chosen ace of spades, in fact three more aces are remaining of which two more aces should be chosen the the probability is $$P=dfracbinom32binom512$$no complexity or conditional probability is needed at all.






share|cite|improve this answer





















  • Thanks, yeah i was forced to use conditional in this problem, but your answer helped me check my answer is correct ! thanks
    – Sacha E
    Jul 17 at 19:36






  • 1




    You're welcome!!
    – Mostafa Ayaz
    Jul 17 at 19:36












up vote
3
down vote



accepted







up vote
3
down vote



accepted






Since you have chosen ace of spades, in fact three more aces are remaining of which two more aces should be chosen the the probability is $$P=dfracbinom32binom512$$no complexity or conditional probability is needed at all.






share|cite|improve this answer













Since you have chosen ace of spades, in fact three more aces are remaining of which two more aces should be chosen the the probability is $$P=dfracbinom32binom512$$no complexity or conditional probability is needed at all.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 17 at 19:25









Mostafa Ayaz

8,6023630




8,6023630











  • Thanks, yeah i was forced to use conditional in this problem, but your answer helped me check my answer is correct ! thanks
    – Sacha E
    Jul 17 at 19:36






  • 1




    You're welcome!!
    – Mostafa Ayaz
    Jul 17 at 19:36
















  • Thanks, yeah i was forced to use conditional in this problem, but your answer helped me check my answer is correct ! thanks
    – Sacha E
    Jul 17 at 19:36






  • 1




    You're welcome!!
    – Mostafa Ayaz
    Jul 17 at 19:36















Thanks, yeah i was forced to use conditional in this problem, but your answer helped me check my answer is correct ! thanks
– Sacha E
Jul 17 at 19:36




Thanks, yeah i was forced to use conditional in this problem, but your answer helped me check my answer is correct ! thanks
– Sacha E
Jul 17 at 19:36




1




1




You're welcome!!
– Mostafa Ayaz
Jul 17 at 19:36




You're welcome!!
– Mostafa Ayaz
Jul 17 at 19:36












 

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