A consignment of 15 record players contains 4 defectives. The record players are selected at random, one by one and examined. The one examined is not put back. Then the probability that the 9th one examined is the last defective is?


I would like to know what is wrong with this way of solving the problem:

Choose 8 first, 3 from defective, rest from working: $^4C_3 \times {{^{11}}C_5}$

Then choose 1, so total number of ways is: $^4C_3 \times {{^{11}}C_5}\times 1$

And total number of ways is ${^{15}C_9}$

So $P(E) = \dfrac{^4C_3 \times {{^{11}}C_5}}{{^{15}C_9}}$


Your count in the denominator counts the first $9$ choices as indistinguishable; we choose $9$ record players, and don't bother tracking the order.

Your count in the numerator, on the other hand, distinguishes the $9$th choice as something special; we choose $8$ players without tracking the order, and then mark the $9$th as special.

In order for the division to make sense, we need to make the same choices about distinguishing order between the numerator and denominator. The easiest way to do that would be to change the rule in the denominator - choose $8$, then choose the $9$th, for $\binom{15}{8}\cdot \binom{7}{1}$ ways.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.