Find the probability of drawing the given cards from a standard deck of 52 cards. (a) with the replacement and b without the replacement
A club, then a diamond
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OpenStudy (ammarah):
@douglaswinslowcooper
OpenStudy (ammarah):
@tkhunny
OpenStudy (ammarah):
is it 4/52 times 13/52? idk how to do this
ganeshie8 (ganeshie8):
how many clubs are there in a standard 52 card deck ?
OpenStudy (ammarah):
4?
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ganeshie8 (ganeshie8):
nope, google a bit
ganeshie8 (ganeshie8):
52 cards --- 26 reds and 26 blacks
OpenStudy (ammarah):
13
ganeshie8 (ganeshie8):
yes !
ganeshie8 (ganeshie8):
so, probability for drawing the "club" first time = 13/52
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OpenStudy (ammarah):
then a diamond so i multiply 4/52?
ganeshie8 (ganeshie8):
since u have taken out one card, only 51 cards will be remaining in the deck
ganeshie8 (ganeshie8):
btw how many diamonds are there in a standard deck ?
OpenStudy (ammarah):
13/51?
ganeshie8 (ganeshie8):
yup !
13/51 is the probability for drawing a diamond second time
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ganeshie8 (ganeshie8):
so the joint probability wud be :-
13/52 * 13/51
OpenStudy (ammarah):
13/204 how about part b??
ganeshie8 (ganeshie8):
wat we did just now is part b
OpenStudy (ammarah):
then what is part a?
ganeshie8 (ganeshie8):
part a) :
since we're replacing the card back into the deck, the second time u will be having 52 cards :-
13/52 * 13/52
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OpenStudy (ammarah):
ohh so without replace ment it will go down
OpenStudy (ammarah):
how about for a jack, then a 7?
OpenStudy (ammarah):
a. 4/52 times 4/52
OpenStudy (ammarah):
b. 4/52 times 4/51
ganeshie8 (ganeshie8):
\(\large \color{red}{\checkmark}\)
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