Using conditional probability$P(A|B) = P(AB)/P(B)$
$P(A) =$ cards space aces
$P(B) =$ ace the spade chosen
I ended up in ~ $P(AB) = \dfrac^3C_1^52C_3$ $3$ methods = Spade,club,heart, spade,heart,diamond and spade society diamond.
You are watching: Two cards are randomly chosen without replacement
$P(B) = (^51C_2)/(^52C_3)$ i think ace that spade chosen, chose another 2 cards
$P(A|B) = <^3C_1/^52C_3)>/<^51C_2)/(^52C3_)>$
I have looked roughly for this question however could only find with 2 cards not 3.Not certain if this is the correct method ?
edited Jul 17 "18 at 20:15
2,69922 yellow badges1616 silver badges2424 bronze title
request Jul 17 "18 in ~ 18:49
17711 silver- badge1111 bronze badges
include a comment |
1 answer 1
energetic oldest Votes
Since you have chosen ace that spades, in fact three much more aces are continuing to be of which two more aces must be preferred the the probability is $$P=\dfrac\binom32\binom512$$no complexity or conditional probability is needed at all.
answer Jul 17 "18 in ~ 19:25
Mostafa AyazMostafa Ayaz
27.8k55 yellow badges1818 silver badges5252 bronze badges
add a comment |
Thanks for contributing an answer to tasiilaq.netematics ridge Exchange!Please be sure to answer the question. Carry out details and share her research!
But avoid …Asking because that help, clarification, or responding to various other answers.Making statements based on opinion; back them up with recommendations or personal experience.
Use tasiilaq.netJax to format equations. Tasiilaq.netJax reference.
To discover more, see our advice on writing great answers.
Sign increase or log in
sign up utilizing Google
authorize up utilizing Facebook
authorize up making use of Email and Password
Post together a guest
email Required, however never shown
Post together a guest
Required, yet never shown
article Your answer Discard
Not the price you're spring for? Browse other questions tagged probability proof-verification or questioning your very own question.
Featured ~ above Meta
Four cards space randomly liked without instead of from an plain deck the 52 playing cards.
See more: Can Fake Pokemon Ds Games Trade ? Do Fake Pokemon Games Work
Probability of acquiring 3 cards in the exact same suit from a deck
discover the probability of illustration 3 aces at arbitrarily from a deck that 52 simple cards if the cards are not replaced.
If five cards are drawn randomly indigenous an plain deck, what is the probability of illustration exactly three challenge cards?
two cards are drawn successively native a deck that 52 cards.
expect we draw two cards there is no replacement the end of a conventional deck that 52 cards
four cards space randomly liked without instead of from an plain deck of 52 playing cards.
Probability selecting three cards the end of a deck
hot Network inquiries an ext hot questions
i ordered it to RSS
concern feed To subscribe to this RSS feed, copy and also paste this URL right into your RSS reader.
stack Exchange Network
site architecture / logo © 2021 ridge Exchange Inc; user contributions license is granted under cc by-sa. Rev2021.9.17.40238