Any STATISTICS (nCr) Pro here ??

[nafu]

hey guys i need some help on Statistics regarding Probability "nCr combination"

there are two question i am really stucked on it from two days. Yes i am Noob, i am really not good in Statistics specifically "probability"

here are two Probability questions...

Probability 1: There are 6 white, 5 Red and 4 Black balls in a Bag. 4 balls are drawn. Find probability that at least two are red balls. Answer is 37/91

Probability 2: Bag 1 contains, 4 white, 2 black balls and bag 2 contains 3 white and 5 blackballs. If 1 ball is drawn from each bag Find the probability that
> Both are white
> Both are black
> One is white and one is black
Answer is 65/176

I need working solution on above two questions and definitely i need quick and better response on it. please don't tell me to Google it as i already done that but could not find the appropriate answer for it.

so if anybody from you have a good grip on it then please post the working solution.

It is really gonna appreciated.

Thanx

 

[hasnain1sds]

Q.1

Ans. [(5C2x10C2)+(5C3x10C1)+(5C4x10C0)]/15C4

Three situations meet the criteria of atleast 2 reds: 2 reds and 2 from remaining 10, 3 reds and 1 from remaining 10 and finally all four reds leaving none to be picked from 10 balls. For probability sum of combinations of all situations divided by total unrestricted outcomes i.e 15C4.

Q.2

Ans. I can't remember how to derive probabilyt using combination in this question but I think it can be solved directly using probab(ility formula. Also you have only given one answer whereas there are three different parts of the question.

i) (4/6)(3/8)

ii) (2/6)(5/8)

ii) [(4/6)(5/8)+(2/6)(3/8)]

 

[nafu]

Q.1

Ans. [(5C2x10C2)+(5C3x10C1)+(5C4x10C0)]/15C4

Three situations meet the criteria of atleast 2 reds: 2 reds and 2 from remaining 10, 3 reds and 1 from remaining 10 and finally all four reds leaving none to be picked from 10 balls. For probability sum of combinations of all situations divided by total unrestricted outcomes i.e 15C4.

Q.2

Ans. I can't remember how to derive probabilyt using combination in this question but I think it can be solved directly using probab(ility formula. Also you have only given one answer whereas there are three different parts of the question.

i) (4/6)(3/8)

ii) (2/6)(5/8)

ii) [(4/6)(5/8)+(2/6)(3/8)]

thanx for the quick response. it is well appreciated.
the First Probability is exactly 100% perfect.

But for Question 2

The book is showing the Answer is one form which 65/176. and from your working solution of Question 2, this answer is not coming. i got 48/192 if adding totalled of all three cases.

 

[hasnain1sds]

thanx for the quick response. it is well appreciated.
the First Probability is exactly 100% perfect.

But for Question 2

The book is showing the Answer is one form which 65/176. and from your working solution of Question 2, this answer is not coming. i got 48/192 if adding totalled of all three cases.

There can't be one answer for all three parts. And if you add the probability of all three parts then the answer should be equal to one as the only possible outcomes are WW,BB and WB or BW. Which book are you using BTW>

 

[nafu]

Business Statistics and Mathematics of B.COM PArt 1 from Muhammad Abdullah

 

[hasnain1sds]

I think its a misprinting then.

 

[nafu]

I think its a misprinting then.

probably

but thanx for providing the solution, Greatly appreciated