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Mathematics 4 Online
OpenStudy (anonymous):

A single card is chosen at random from a shuffled standard fifty-two card deck (no jokers). Find the following probabilities. 1)What is the probability that the card is black? 2)What is the probability that the card is a heart? 3)What is the probability that the card is either a diamond or a 4? 4)What is the probability that the card is at least a 7 and at most a 9? That is, P(≥7∩≤9)

OpenStudy (anonymous):

I guess 1) 1/2 2) 1/4 3) 17/52 or 16/52 i don't know for sure 4)21/52

OpenStudy (amistre64):

if we out all the diamonds and 4s into a pile, how many cards would be in the pile?

OpenStudy (amistre64):

if w put all the 7s, 8s and 9s in a pile, how many would be in the pile?

OpenStudy (amistre64):

21 might be a typo ... 12/52 seems better for that one

OpenStudy (anonymous):

k

OpenStudy (anonymous):

im a bit confuse 12/52 is from which question?

OpenStudy (amistre64):

the one that has 21 in it ...

OpenStudy (anonymous):

question 4) ?

OpenStudy (anonymous):

\[P \left( A \cup B \right)=P \left( A \right)+P \left( B \right)\]-P(A intersection B)

OpenStudy (anonymous):

all are correct except for 4 ?

OpenStudy (amistre64):

if we put all the diamonds and 4s into a pile, how many cards would be in the pile?

OpenStudy (anonymous):

there are 13 diamond and 4 (club diamond, heart and spade)

OpenStudy (amistre64):

13 diamonds, plus 1 card from each of the other 3 suits 13+3

OpenStudy (anonymous):

16/52

OpenStudy (amistre64):

yes

OpenStudy (anonymous):

ok

OpenStudy (amistre64):

if we take a 7, 8, and 9 from each suit ... so 3 cards out of 4 suits would be how many cards?

OpenStudy (anonymous):

12 ?

OpenStudy (amistre64):

yes, so 12/52 for that last one .. not 21/52

OpenStudy (anonymous):

ok i see

OpenStudy (anonymous):

now we are done here ?

OpenStudy (amistre64):

the rest are fine, so yeah

OpenStudy (anonymous):

thank you so much for the help

OpenStudy (amistre64):

good luck :)

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