Question

A bag contain 10. identical balls of which 4 are
black and 6white a random sample of size 4 is
drawn without replacement what is the
conditional probability that the sample contain
exactly two black given that it contain at least 1
white ball..

Please show step plz

Answer 1

this is binomial...with a twist

P(b) = .4 , P(w) = .6

(b + w)^4

b^4 + 4 b^3 w + 6 b^2 w^2 + 4 b w^3 + w^4

P(w ≠ 0) = 1 - P(b = 4) = 1 - .4^4

P(b = 2) = 6 * .4^2 * .6^2

divide to find the conditional

Answer 2

i dont know what to do

Answer 3

I honestly dont know what you are working to. It seems to me you are looking for 2blacks, 1 white out of 9 balls.
you can do that
bbw
wbb
bwb three ways.
Pr(2black,1w given 1W)=3*4/9*3/8*5/7 =12*15*1/(9*8*7)=4*5/8*7
= 3/14 check my math, I did it in my head.

Honestly, I am not certain what you mean by "given"

Answer 4

well using C(n,r)

for 2 black
p(b) =.4
p(not black) = .6

n = 10
r = 2

P(10,2) = C(10,2) .4^2 .6^8
C(10,2) = 10!/[8! 2!]
= 10*9/2 = 45
so
45 * .16 * .0168
=.121

now for no whites, that is P of 4 blacks
6/10 * 5/9 * 4/8 * 3/7 = .0714
1-.0714 = .929 = probability of at least one white
so I get
.121 / .929

Answer 5

I get 5/14 = 0.3571

Damon's answer above is

P(b=2)/P(w≠0) = 0.3546

Seems they ought to be the same ...

Answer 6

I got .121/.929 = 0.130

Answer 7

"you can lead a horse to water...but you can't make him drink"

Answer 8

LOL use Scott's answer. I used 10 for number of trials, did not read correctly.

Answer 9

sir scott please show full work sir plz thanks

Answer 10

He did, right here:
P(w ≠ 0) = 1 - P(b = 4) = 1 - .4^4

P(b = 2) = 6 * .4^2 * .6^2
===============================

prob of some white = 1-.4^2
= .84
the prob of 2 black = 6*.4^2*.6^2
= .3456

divide
.84/.3456

so

Answer 11

.3456/.84 = .4114