Question

A bag contains 6 white and 8 black balls.If 4 balls are drawn.What is the probability that:
a)All are black
b)2 are white and 2 are black

Answer 1

@nincompoop

Answer 2

Sooo you have 14 balls.
a) You have to pick all black:
(8/14)*(7/13)*(6/12)*(5/11)

Answer 3

Got it,and after this? :)

Answer 4

@bahrom7893

Answer 5

b) 2 are white and 2 are black:
(6/14)*(5/13)*(8/12)*(7/11)

Answer 6

basically probability of picking a ball of a certain color:
number of balls of that color / number of balls total

Answer 7

@bahrom7893 it never said you necessarily draw 2 white and then 2 black!

Answer 8

I don't get the b part.

Answer 9

@oldrin.bataku is right. you have to multiply my answer for b by 4!

Answer 10

but wait two of them are the same color, so 4!/2!

Answer 11

I meant to say 2 pairs are the same color

Answer 12

Okay now can you solve and show the first part? That is part a.

wait we have to write it like this:

\[\frac{ 6 }{ 14 }\frac{ 5 }{ 13 }\frac{ 8 }{ 12 }\frac{ 7 }{ 11 }\]

Answer 13

Multiply sign in between?

Answer 14

how come I didn't get a notification?

Answer 15

yea because basically we have to keep getting black balls:
8 black balls out of 14 total = 8/14 (one black ball is gone)
7 black balls out of 13 total = 7/13, (7 black balls are remaining, and there are 13 in total) etc. follow the pattern

Answer 16

I swear to god i don't get it.

Answer 17

I miss...@terenzreignz Now..

Answer 18

@terenzreignz

Answer 19

You can use the hypergeometric distribution

Answer 20

Let X = # of black balls
14 balls total
You choose 4 of them

If you want all of them to be black, then plug in x = 4

\[f(x)=\frac{\left(\begin{matrix}8 \\ x\end{matrix}\right) \left(\begin{matrix}6 \\ 4-x\end{matrix}\right)}{\left(\begin{matrix}14 \\ 4\end{matrix}\right)}\]

Answer 21

b) you can just apply the same logic but you consider the case for 2 black balls (put in x=2) then just multiply that with the result for white=2 ... you can re-write f(x) for white balls and compute it for that,

Answer 22

@kirbykirby is correct; hypergeometric distribution is for where we don't have replacement