Box containing Red and black balls

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tl;dr

A box containing 4 red balls and 6 black balls . Three balls are selected randomly from the box one after the another, without replacement . The probality that the selected set has one red ball and two black balls is

Do you think order matters? No. Why? Because if we select $RBB$ or $BRB$ or $BBR$ any of them occuring is favorable, and we are done, you might as well consider order but that would cancel eventually. So total possibilities in three turns: $$\binom{10}3$$ Ways to do favorable event: $$\binom62\binom41$$ Probability: $$P=\frac{\binom62\binom41}{\binom{10}3}=\frac12=50\%$$

Now let me consider order:

Total ways: $$10*9*8$$ Favourable ways: $$RBB:\binom41\binom61\binom51\\ BRB:\binom61\binom41\binom51\\ BBR:\binom61\binom51\binom41$$ Or simply no. of ways of arranging $B,R,B$ times "you know": $$\frac{3!}{2!}6*5*4=6*5*4*3$$ Probability is: $$\frac{6*5*4*3}{10*9*8}=\frac12=50\%$$

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jayagopi jayamani

Updated on August 01, 2022

Comments

jayagopi jayamani over 1 year
A box containing 4 red balls and 6 black balls . Three balls are selected randomly from the box one after the another, without replacement . The probality that the selected set has one red ball and two black balls is

I have a doubt in this problem .

There are 10 balls

Sample Space = First i pick 1st ball X then 2nd ball X then 3rd ball

So Sample Space = 10 * 9 * 8

Ist trial I pick 1 red ball from 4 red balls = 4C1 2nd trial I pick 1 black ball from 6 black balls = 6C1 3rd trial I pick 1 black ball from 5 black balls = 5C1

a.) Probality = (4C1*6C1*5C1)/(10*9*8)
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        OR
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b.) I pick 1 red ball from 4 red balls =4C1 I pick 2 black balls from 6 red balls = 6C2

Probality = (4C1*6C2)/(10*9*8)

Is it a or b ? Pleas elaborate ???