Question

3 cards are drawn from a standard deck of 52 cards. Find the probability of drawing either 3 red cards or 3 face cards.

Answer 1

There are 12 face cards in a standard 52 card deck are 12/52=3/13

And 26 red cards -13 hearts and 13 diamonds.
26/52=1/2

Answer 2

When we multiply both the probabilities it comes as 3/26 so it's 6/52.

Answer 3

none of the answers i got

Answer 4

P(red) = 1/2
P(face) = 4/13, or 3/13 .... depends on how you count an ace

but some face cards are red .... so we have to adjust for that

Answer 5

Probability (of drawing 3 red cards) 26/52
Probability (of drawing 3 face cards) 12/52
Probability (of drawing a face card and bring red) = 6/52

P (A or B) = P(A) + P(B) - P ( A and B)
= 26/52 + 12/52 - 6/52 =

Answer 6

I was almost there :D

Answer 7

might want to adjust for "not replacing" maybe?

Answer 8

Yes that is what I wanted to ask?

Answer 9

Than you have to use a hyper geometric distribution...

Answer 10

if we divvy it up int: r = red not face, R = red face, f = face not red .... we could brute it like this:

rrr

fff

rrR
rRr
rRr
rRR
RrR
RRr

ffR
fRf
Rff
fRR
RfR
RRf

RRR

Answer 11

P(red) = (26C1/52C1) x (25C1/51C1) x (24C1/50C1) =

p(face) = (12C1/52C1) x(11C1/51C1) x (10C1/50C1) =

p(face and red) = (6C1/52C1) x (5C1/51C1) x (4C1/50C1)) =

Could you do it like this too @amistre64

Answer 12

a.)84/137
b.)168/663
c.28/221
d.)120/1326

Answer 13

Its C. just do what I did

P(red) + P(face) - P(both)
2/17 + 11/1105 - 1/1105 = ? simplify..

Answer 14

curious if my brute method pays out :)

if we count aces as numbers: 20 red not face
6 red face
6 face not red

((20*19*18)+(6*5*4)+(6*5*4)+6(20*19*6)+6(6*5*6))/(52*51*50) = 14/85, thnx the wolf

Answer 15

if we count aces as faces: 18 red not face
8 red face
8 face not red

((18*17*16)+2(8*7*6)+6(18*17*8)+6(8*7*8))/(52*51*50) = 956/5525 ... prolly not that way then

Answer 16

slight error in my haste lol ... a counted rrR the same as RRr

Answer 17

rrr (20*19*18)

fff (6*5*4)

rrR
rRr 3(20*19*6)
Rrr

rRR
RrR 3(20*6*5)
RRr

ffR
fRf 3(6*5*6)
Rff

fRR
RfR 3(6*5*6)
RRf

RRR (6*5*4)

all over (52*51*50)

Answer 18

28/221 yep

http://www.wolframalpha.com/input/?i=%28%2820*19*18%29%2B2%286*5*4%29%2B3%2820*19*6%29%2B3%2820*6*5%29%2B6%286*5*6%29%29%2F%2852*51*50%29