Question

Probability,
A bag contains 4 white, 3 black, 2 red balls . Balls are drawn from the bag one by one without replacement. The probability that 4th ball is red is equal to...

Answer 1

We can pick red ball in the fourth draw in two ways :
1) no red ball was picked in the first 3 draws
2) only one red ball was picked in the first 3 draws

Answer 2

i am able to formulate all the favorable cases but not all the cases

Answer 3

All the cases is simply 9P4 = 9*8*7*6 right ?

Answer 4

may be... um confused..

Answer 5

(assume all the balls are distinguishable)

Answer 6

1) no red ball was picked in the first 3 draws

the probability for this case is
`7/9*6/8*5/7*2/6`

yes ?

Answer 7

thats what bothering me that my favorable cases are altered by my sample space
and how can we assume all the balls are indistinguishable, thats gonna affect cases

Answer 8

it wont affect the final probability as we're dividing by assuming the "same" thing through out

Answer 9

can you get a final answer?

Answer 10

1) no red ball was picked in the first 3 draws

the probability for this case is
`7/9*6/8*5/7*2/6`


2) only one red ball was picked in the first 3 draws
the probability for this case is
`3(2/9*7/8*6/7)*1/6`


final probability = `7/9*6/8*5/7*2/6` + `3(2/9*7/8*6/7)*1/6`

Answer 11

https://www.wolframalpha.com/input/?i=+7%2F9*6%2F8*5%2F7*2%2F6+%2B+3%282%2F9*7%2F8*6%2F7%29*1%2F6

Answer 12

its correct.. please explain...i understood the two cases rest after um blank

Answer 13

i got it thanks :)

Answer 14

okay :)

Answer 15

theres a second part also
The probability that 5th ball is black given that the 4th ball is red is equal to

Answer 16

lets define below two events

x : "5th ball is black"
y : "4th ball is red"

we need to find P(x|y)

Answer 17

P(x|y) = P(x and y)/P(y)

Answer 18

we already know P(y) = 2/9 from part a

Answer 19

just need to find P(x and y)

Answer 20

wid me so far ? :)

Answer 21

yeah.. ;)

Answer 22

part b looks hard
is the answer 43/56 ?

Answer 23

nope

Answer 24

33/70 ?

Answer 25

nope

Answer 26

1/12 ?

Answer 27

nahh

Answer 28

1/8, 3/8,2/11,3/11 are the options

Answer 29

oh good u are given options

Answer 30

was thinking you're entering the answers into a grader to check

Answer 31

um solving papers with basic patterns

Answer 32

ok i forgot to divide by 2/9 earlier
the answer is (1/12)/(2/9)

Answer 33

i found a really simple way to work both parts

Answer 34

do tell

Answer 35

first we translate the given problem into a well known problem :

forming a 9 digit number without repitition

Answer 36

`A bag contains 4 white, 3 black, 2 red balls`

assign numbers to colors like this
1, 2, 3, 4 : white
5, 6, 7 : black
8, 9 : red

Answer 37

then part a of the problem translates like this :

Using the digits {1,2,3,4,5,6,7,8,9}, find the number of 9 digit numbers that can be formed such that the 4th digit is 8 or 9

Answer 38

i got the second part too... you are incredible!!

Answer 39

may i see how u worked both parts xD

Answer 40

yeah 4th place can be filled with 2 balls out of 9 balls so 2/9
and is the 4th place is already red, remaining ball are 8 and 3 black can occupy the 5th place so 3/8
right?

Answer 41

Wow! thats amazing!

Answer 42

that didn't occur to me, looks really neat

Answer 43

haha thanks :)
up for a geometry question? :P

Answer 44

sure i prefer geometry to probability yeah ;p

Answer 45

@thomas5267 has something to say i think

Answer 46

I will say it when I am ready lol. Go look for other questions!

Answer 47

I used computer to brute force all permutations of ball of length 4 and count the permutations with red ball on position 4. There is 71 such permutations and 20 of them have a red ball on position 4. Did I do something wrong?