Question

Bob is moving and all of his sports cards are mixed up in a box. Twelve
cards are baseball, eight are football, and five are basketball. If he reaches
in the box and selects them at random, find each probability.
1. P(3 football) 2. P(3 baseball)
1) 8C3/25C3=56/2300
1) exactly the same 1) ?

Answer 1

my answer 1 and 2 ?

Answer 2

how many cards are being drawn...is it without replacement?

Answer 3

It's an 8/25 for the first football card..

Answer 4

25 total, drawn 3

Answer 5

hmmm....;)

Answer 6

if this is without replacement then the probability follows a hypergeometric distribution

Answer 7

(8/25)(7/24)(6/23) right?

Answer 8

yes ... which is the same as 8C3/25C3

Answer 9

what do you mean 8C3/25C3

Answer 10

8C3/25C3 yes correct ,that I did on frist one ,
secont one ?

Answer 11

how many baseball cards are there ;)

Answer 12

lol grrr.

Answer 13

you mean football cards!? lol

Answer 14

5

Answer 15

2. P(3 baseball)

"Twelve cards are baseball"

Answer 16

12C3/25C3 ?

Answer 17

yes

Answer 18

I need more help Please!
P(1 basketball, 2 football)

Answer 19

ok...what is the denominator?

Answer 20

25

Answer 21

just 25?

Answer 22

what was it in the other two problem when you took 3 cards?

Answer 23

took 3 than 2 left

Answer 24

no...the overall experiment is that you are picking 3 cards...how many ways can you do this?

Answer 25

3!

Answer 26

how many ways can I choose 3 cards from 25

Answer 27

25!/(25-3)!3!

Answer 28

yes...which is 25C3

Answer 29

that is our denominator

Answer 30

P(1 basketball, 2 football)

lets break this up into two parts

Answer 31

how many ways can we choose 1 basketball card from the 5 we have?

Answer 32

5!/(5-1)!1!

Answer 33

which is 5C1

good

now

how many ways can we choose 2 football cards from the 8 we have?

Answer 34

8!/(8-2)!2!

Answer 35

yes 8C2

multiply those two numbers we found and that is your numerator

Answer 36

you get it?

Answer 37

\[\frac{_5C_1\cdot _8C_2}{_{25}C_3}\]

Answer 38

yes,how about : P(2 basketball, 1 baseball)

2 basketball = 5C2
1 baseball = 12C1

Answer 39

yes

Answer 40

what denominator?

Answer 41

\[\frac{_5C_2\cdot _{12}C_1}{_{25}C_3}\]

Answer 42

the same

Answer 43

you are performing the same experiment over and over....taking 3 cards

Answer 44

I don't get is 25C3, I under stand 25 is total ,
3?. How you get 3

Answer 45

you are taking 3 cards

Answer 46

3 card from the frist one ?

Answer 47

prob=(number of ways to get a particular result from an experiment)/(number of ways to do the experiment)

Answer 48

you wanted 2 basketball, 1 baseball

that is 3 cards

Answer 49

P(2 basketball, 1 baseball)
deominator is 25C3

P(1 football, 2 baseball)
deominator is 25C3

P(1 basketball, 1 football, 1 baseball)
deominator is 25C3

P(2 baseball, 2 basketball)
deominator is 25C4

P(2 football, 1 hockey)
deominator is 25C3

Answer 50

if that is all the info you are given then yes...those are correct

Answer 51

thank, great help