Question

I need help please (this is due by tommorrow)
A bag contains five red chips, three white chips, and two blue chips. Three chips are
to be selected at random, without replacement. Find the probability that
a. all are red
b. the first two are red and the third is blue
c. the first is red, the second is white and the third is blue

Answer 1

this is not nearly as hard is it seems, as long as you know how to compute 10 choose 3 and stuff like that
you know how to do that?

Answer 2

um no i dont

Answer 3

oh nvm now that i actually read the question, you don't need that at all

Answer 4

lets do the first one
what is the probability that the first one selected is red?

Answer 5

5/10 wich is reduced to 1/2

Answer 6

ok now we have a red one in our hand (because we are not replacing it)
what is the probability that the next one chosen is also red?

Answer 7

4/10

Answer 8

not quite

Answer 9

there are 4 red ones left for sure
but how many chips are left in the bag?

Answer 10

thers 9 left in the bag

Answer 11

zactly
would you like to change your answer from \(\frac{4}{10}\) ?

Answer 12

if thats the correct answer i would change it

Answer 13

no 4/10 was incorrect
since there are 4 red ones left out of the total of 9, the probability that the next one is red is \(\frac{4}{9}\)

Answer 14

one more to go
we took out 2 red chips
how many red chips are left, how many chips total?

Answer 15

3 left, 7 total

Answer 16

i think 8 total since you started with 10 have have taken out two

Answer 17

take all these probabilities and multiply them together to the get the probability of all three being red

Answer 18

oh yeah thats right

Answer 19

i.e.
\[\frac{5}{10}\times \frac{4}{9}\times \frac{3}{8}\] cancel first, multiply last

Answer 20

60/720

Answer 21

wait im confused what do you cancel and multiply last

Answer 22

you already said \(\frac{5}{10}=\frac{1}{2}\) right?

Answer 23

yes

Answer 24

just got logged out

Answer 25

\[\frac{5}{10}\times \frac{4}{9}\times \frac{3}{8}\] every number in the numerator cancels with some number in the denominator
your numerator will be 1 when you are finished cancelling
hence "cancel first, multiply last"

Answer 26

yes i got 1 for the numerator and 12 for the denominator

Answer 27

good

Answer 28

so would my answer be 1/12

Answer 29

yes

Answer 30

okay

Answer 31

others work exactly the same way

Answer 32

can you help me on the last one

Answer 33

i find the last one more difficult

Answer 34

no not at all
c. the first is red, the second is white and the third is blue

Answer 35

what is the probability the first is red?

Answer 36

4/9

Answer 37

?

Answer 38

so 4/10 since they're asking for one red

Answer 39

i think you may be confused by the question
we are starting over

Answer 40

A bag contains five red chips, three white chips, and two blue chips. Three chips are to be selected at random, without replacement. Find the probability that the first is red, the second is white and the third is blue

Answer 41

so once again, what is the probability that the first one is red?

Answer 42

1/10

Answer 43

hmm no
same as last time

Answer 44

five out of ten are red

Answer 45

ok i got that part

Answer 46

now one red one is gone
what is the probability that the next one is white?

Answer 47

3/9

Answer 48

yup
one more and you are done

Answer 49

one red one is gone
one white one is gone
what is the probability that the next one is blue?

Answer 50

the next one would be 2/8

Answer 51

so it would
final job is
\[\frac{5}{10}\times \frac{3}{9}\times \frac{2}{8}\] cancel first, multiply last

Answer 52

ok i got 1/24 for my final answer