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In a box containing $15$ apples, $6$ apples are rotten. Each day one apple is taken out from the box. What is the probability that after four days there are exactly $8$ apples in the box that are not rotten?

Answer to this question is $12/91$.

I proceeded this question in this way: $$n(S)=15 \cdot 14 \cdot 13 \cdot 12$$ Now to calculate $n(E)$:

We must choose $1$ out of $9$ good apples(which are not rotten) and $3$ out of $6$ rotten ones. So a sample sequence can be $GRRR$ where $R$=rotten and $G$=good. This sequence can be arranged in $4$ ways, so total $$n(E)= \binom{9}{1} \cdot \binom{6}{3} \cdot 4$$ so my answer is coming $2/91$(there is a extra six in my denominator).

Answer in the book is something like this $$4\left(\frac{9}{15}\right)\left(\frac{6}{14}\right)\left(\frac{5}{13}\right)\left(\frac{4}{13}\right)$$

I am able to understand this solution.

My questions:

1) $n(S)=15 \cdot 14 \cdot 13 \cdot 12$ is this the complete sample space shouldn't we multiply it with number of different arrangement(like $RRRR,RRGG,GGGG,RGRG, \ldots$)

2) How is my solution different with book's solution?

3) Suppose we want to draw four balls. Is there any difference between drawing four balls at a time and drawing one ball at a time please help!

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  • $\begingroup$ Please see this tutorial on how to typeset mathematics on this site. You can see how I typeset your equations by right clicking on an equation, then selecting Show Math As TeX Commands. $\endgroup$ – N. F. Taussig Jul 11 '15 at 13:50
  • $\begingroup$ Thanks for info @N.F. Taussig $\endgroup$ – blue boy Jul 11 '15 at 18:45
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For a hypergeometric problem in which the order of events is unspecified, one way is what the book has used, viz. direct multiplication of probabilities with a multiplication factor of 4 to take care of possible order of events.

The other way is to use combinations. You have mixed up the two.

Using combinations, the formula will simply be $\dfrac{{9\choose 1}\cdot{6\choose 3}}{15\choose 4}$

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  • $\begingroup$ Thanks...I have one doubt that..what about arrangement say for example......when you choose (15c4) the four selected can be arranged in different ways...one more thing is that (this is really silly) all the rotten or good apples are not alike ( if i am not mistaken) so an element from a sample space will look like (R1)(R2)(R3)(G1) rather than RRRG. $\endgroup$ – blue boy Jul 11 '15 at 12:13
  • $\begingroup$ I know standard problem related to hypergeometrical distribution are solved in this way(Thanks again for helping @true blue anil)....just reinventing the wheel ....see if it is right or wrong.....................................................answer=4(9c1)(6c3)3!/(15c4)4!=(9c1)(6c3)/(15c4) $\endgroup$ – blue boy Jul 11 '15 at 12:30
  • $\begingroup$ We need to keep it simple: just 2 categories: good and bad ! $\endgroup$ – true blue anil Jul 11 '15 at 12:38
  • $\begingroup$ You're welcome ! $\endgroup$ – true blue anil Jul 12 '15 at 2:15
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$$\frac{\binom91\binom63}{\binom{15}4}=\frac{12}{91}=4\frac9{15}\frac6{14}\frac5{13}\frac4{12}$$

As you see the two different approaches lead to the same result.

Picking out $1$ good and $3$ rotten apples can be done in $\binom91\binom63$ ways. The factor $4$ that you use in calculating $n(E)$ is wrong.

There is no essential difference between drawing $4$ balls at a time or $4$ balls one by one without putting them back.

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