# probability of selecting dvds

studying for a test, cant figure out this probability problem

A bin at Blockbuster contains 100 DVD's of which 20 are defective. You randomly select 10 and try them out at home. You discover that there 2 defective DVD's in the 10 that you selected. The store now allows you to select 2 replacements from the same bin (which now only has 90 DVD's in it, since you already removed 10). What is the probability that none of 10 DVD's you fi nally end up with are defective? What is the answer to this question if you fi nd k defectives in your initial choice?

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HINT: You’re selecting $2$ replacements from a bin that has $90$ DVDs, of which $18$ are defective (since you removed $2$ defective DVDs in your first selection of $10$). That means that the bin contains $72$ good DVDs. You need to select $2$ good DVDs in order to end up with $10$ good ones.

There are $\dbinom{90}2$ ways to choose $2$ DVDs from the bin, and $\dbinom{72}2$ ways to select $2$ good DVDs.

• What do you do with these numbers to get the desired probability?

• What changes in the calculation if $2$ is replaced by $k$?

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you multiply them to get the total number of good dvds? –  dotdotdot Feb 14 '13 at 4:15
@dotdotdot: You don’t want a number of DVDs: you want the probability that both of the replacement DVDs are good. Basic fact: if something can happen in $n$ equally likely ways, and $k$ of those ways are good, then the probability of a good outcome is $\frac{k}n$. So what should you do with $\binom{90}2$ and $\binom{72}2$? –  Brian M. Scott Feb 14 '13 at 4:19
divide 90C2/72C2 –  dotdotdot Feb 14 '13 at 4:35
@dotdotdot: But that gives you approximately $1.5669$; is that a possible value for a probability of anything? –  Brian M. Scott Feb 14 '13 at 4:36
80 would be non defective the other 20 would be defective right? –  dotdotdot Feb 14 '13 at 7:50