Question

A bag contains 5 red marbles, 6 white marbles, and 5 blue marbles. Find p(red and blue)

Answer 1

the probability is 10/16 reduced to 5/8

Answer 2

^wouldn't that be P(red OR blue)

Answer 3

5/8 is what i got for P(red OR blue)

Answer 4

no because the question says red "AND" blue @kirbykirby

Answer 5

1st draw: getting a red is 5/16
2nd draw: getting a blue is 5/15, since there are 15 marbles left. Then you use the product rule

Answer 6

no.....just no

Answer 7

I'm so confused

Answer 8

Then how would you find P(red and blue and white and red) ?

Answer 9

If you do by adding the individual probabilities, then you will go over 1 and probabilities can't exceed 1

Answer 10

"I'M DONE" Because me and @tmason8 are correct so it's over

Answer 11

your problem doesn't specify whether this is a single draw of two marbles, or two draws, or whether marbles are put back....
please clarify, or the question cannot be answered

Answer 12

huh?

Answer 13

if the marbles are drawn successively and not put back, then @kirbykirby is correct
I have no idea where you got your answer from @gswag98

Answer 14

The question doesn't specify that, which is why I am so confused. I am used to them saying if it is single draw etc.

Answer 15

a single draw is the same as tow successive draws without any marbles being put back. As @kirbykirby stated, P(red)=5/16, then P(blue)=4/15 since once less marble is now in the bag. You can then treat the two events as independent.

Answer 16

typo P(blue)=5/15

Answer 17

Now that I think of it, if it's not successive draws, then we'd have to divide by 2 right? or think of it as
\[ \large \frac{{5 \choose 1}{5 \choose 1}}{{16 \choose 2}}\]

Answer 18

*multiply by 2

Answer 19

no, i don't think it would change your answer if the draws are not successive, because\[P(R\cup B)=P(R)P(B|R)=P(B)P(R|B)\]

Answer 20

the conditional probabilities for red/blue respectively are the same regardless of order, and drawing 'at the same time' can always be represented as successive draws without replacement

Answer 21

What is the mean, variance, and standard deviation of the values? Round to the nearest tenth. 1, 9, 4, 12, 13, 13

Answer 22

please post each question separately

Answer 23

how?

Answer 24

close this post (little blue button near start of thread) then do a new one just like before

Answer 25

Hm but if we draw blue first then red, the order doesn't matter, so are we double counting? @TuringTest

Answer 26

hmm i think you are right

Answer 27

by the total probability theorem, i think\[P(R\cup B)=P(R)(P(B|R)+P(B)P(R|B)\]right? so that makes you right, we need to multiply your original answer by 2 since the terms above are equal
man, i just took probability and i'm already forgetting it :P

Answer 28

yes of course you are right, silly me XD

Answer 29

don't make sense

Answer 30

we went above your level, I think. sorry, but you probably have a set way to do this from your book, but kirby and I are going off general probability laws

Answer 31

ok

Answer 32

Ah yes total probability rule hehe.

Answer 33

@LearningIsAwesome

Answer 34

Can you make a new question and tag me in it? I think I know this

Answer 35

Can you make a new question and tag me in it? I think I know this @tmason8

Answer 36

\[\large \frac{{5 \choose 1}{5 \choose 1}}{{16 \choose 2}} \]
Or think of P(red and blue) as
[P(red then blue) or P(blue then red] = P(red then blue) + P(blue then red)
= (5/16)(5/15) + (5/16)(5/15)

Answer 37

hmmm... why is this still being replied to I mean tired of getting a notification of this problem you know I think me and @tmason8 are right that's all too it

Answer 38

I know that you can add the fractions 5/16 + 5/16 together

Answer 39

can't*

Answer 40

that's for P(red "or" blue)