PROBABILITY

Question

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 am getting this (4/10)*(6/9)*(5/8) is it correct And is order have any significance here.?

Comments

I don`t think so ! 3c1*(4/10)*(6/9)*(5/8)=1/2 ?

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@ saxena0612 Why are you doing selection 3c1 ?

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Because given condition is that set has one red and two black balls LIKE RBB what in what order it has been picked is not mentioned Thus It could be RBB, BBR,BRB i choosed 1?

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$\frac{\binom{4}{1} * \binom{6}{2}}{\binom{10}{3}}$

Answer 1

This problem is an example of hypergeometric distribution..It is similar to binomial distribution but differs with respect to following things :

a)  The population being considered here is finite and the mode of event is "without replacement".

b) The probability changes for each individual event , which violates the assumption of Binomial distribution that probabilty should remain the same.

So here in total we have 10 balls , which is a finite population and out of which we have 4 red and 6 black balls..Now according to the question , favorable event is the one in which we have one red balls and two black balls..

Total number of events    =   Number of ways in which we can pick 3 out of 10 balls    =    ¹⁰C₃

Now number of favorable events  =       Number of ways 1 red ball can be picked out of 4 red balls * Number of ways 2 black balls can be picked from 6 balls           =       ⁴C₁  * ⁶C₂

Hence required probability            =      ( ⁴C₁  * ⁶C₂ )  /  ¹⁰C₃

                                                                  =      ( 4 * 15 * 6 ) / ( 10 * 9 * 8 )

                                                  =      1 / 2     

Ref  :   https://en.wikipedia.org/wiki/Hypergeometric_distribution

Comments

@Habibkhan Am I correct?

In Binomial Distribution the population being considered is finite and the mode of the event is "with replacement".

If the population is very very large then it turns to Poisson distribution.

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Yes u r right.. Further for infinite population whether we have replacement option available or not, it will be Binomial distribution.. But for finite population it matters..

When population is very large and probability of occurrence is very very small then Binomial distribution becomes Poisson distribution..