Ask your own question, for FREE!
Mathematics 8 Online
OpenStudy (anonymous):

a standard deck of card contains 52 cards. one card is selected from the deck. a. p(heart or spade) b.p(heart or spade or diamond) c. p(queen or spade)

OpenStudy (anonymous):

i think a. is 1/4 b.3/4

OpenStudy (anonymous):

but my best guess for c is 17/52 im not sure if that right

OpenStudy (kropot72):

a. The probability of heart or spade is: P(heart) + P(spade) There are 13 cards in each suit, so: P(heart) + P(spade) = 13/52 + 13/52 = you can calculate

OpenStudy (anonymous):

so 1/2

OpenStudy (kropot72):

b. The probability of heart or spade or diamond is: P(heart) + P(spade) + P(diamond) = 13/52 + 13/52 + 13/52 = 3/4

OpenStudy (anonymous):

yay i got that right

OpenStudy (kropot72):

Yes, 1/2 is correct for a.

OpenStudy (anonymous):

for c i keep coming up with 17/52

OpenStudy (anonymous):

=.33

OpenStudy (anonymous):

is c right

OpenStudy (anonymous):

a.) 1/2 b.) 3/4 c.)17/52 or .33

OpenStudy (anonymous):

thank you

OpenStudy (kropot72):

There are 12 cards in the suit of spades without the queen of spades. And there are 4 queens; So: P(queen or spade) = 4/52 + 12/52 = 4/13

OpenStudy (anonymous):

c.) p(queen)+p(spade)=4/52+13/52=17/52 or.33

OpenStudy (anonymous):

oh thank you again i forgot you can count something twice

OpenStudy (kropot72):

You're welcome :)

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!