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A Box contain 5 red , 6 blue, 4 white balls. three ball is drawn at random. what is the probability of getting all the red colored balls?

In the answer we are considering n(E)=$^5C_3$$=$$^5C_2$ how?

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I like to do these in steps.

First step: What is the probability of pulling out a red ball? Clearly it's $\frac{5}{5+6+4} = \frac{5}{15} = \frac{1}{3}$.

Second step: After removing one red ball, what's the probability of pulling out another red ball? In this case it is $\frac{4}{4+6+4} = \frac{4}{14} = \frac{2}{7}$.

What is the last step? What can you do to get the overall probability that these three events happen in succession?

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  • $\begingroup$ actually the answer of 5C2/15C2 is 2/91 $\endgroup$ – Vignesh Vicky Jul 31 '13 at 4:42
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Probability of getting 3 red balls$=$$ ^5C_3/15C_3= (5!/3!*2!)/(15!/3!*12!)= (5*4*3)/(15*14*13)= 2/91$

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  • $\begingroup$ same answer but they considered 5C2 instead of 5C3? $\endgroup$ – Vignesh Vicky Jul 31 '13 at 4:50
  • $\begingroup$ nCx=nCn-x, it is a property. can be proven easily. because in denominator we use both x! and n-x! $\endgroup$ – Ramit Jul 31 '13 at 4:52

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