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From $27$ pieces of luggage, an airline handler damages a random sample of $4$. The probability that exactly one of the damaged pieces of luggage is insured is twice the probability that none of the damaged pieces are insured. Calculate the probability that exactly two of the four damaged pieces are insured.


Hey guys! I tried solving this problem by assuming that there were $18$ insured pieces of luggage and 9 not insured pieces of luggage because of "The probability that exactly one of the damaged pieces of luggage is insured is twice the probability that none of the damaged pieces are insured". Then I did $\frac{\binom{18}{2} \times \binom{9}{2}}{\binom{27}{4}}$. It didn't end up with the right answer. The right answer is $.27$ (rounded). Can you guys please explain to me what I did wrong or if I made a wrong assumption?

Thank you so much!

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2 Answers 2

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Assume that each of the $4$ luggage items damaged by the handler was chosen independently and with equal probability. Let the number of insured items in the lot of $27$ be $i$ and uninsured $n$.

The probability that the handler damages $4$ items, only one of which is insured, is $$ \frac{\displaystyle {i \choose 1}{n\choose 3}}{\displaystyle{27\choose 4}}\,. $$

The probability that the handler damages $4$ items, none of which is insured, is $$ \frac{\displaystyle {n\choose 4}}{\displaystyle{27\choose 4}}\,. $$

We are told that the relationship between these probabilities is $$ \frac{\displaystyle {i \choose 1}{n\choose 3}}{\displaystyle{27\choose 4}}{}={}2\frac{\displaystyle {n\choose 4}}{\displaystyle{27\choose 4}}\,. $$

Simplifying, this implies that $$ n-2i{}={}3\,. $$ But, recall, that we also know $$ i{}+{}n=27\,. $$ Thus, solving this system of equations yields $i{}={}8$ and $n{}={}19$, from which the probability that exactly $2$ of the $4$ damaged items are insured may be calculated as $$ \frac{\displaystyle {8 \choose 2}{19\choose 2}}{\displaystyle{27\choose 4}}{}\approx{}0.2728\,. $$

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  • $\begingroup$ Thank you so much! You were such a big help! $\endgroup$ Dec 20, 2014 at 3:08
  • $\begingroup$ very cool question and answer $\endgroup$ Jul 6, 2019 at 20:53
  • $\begingroup$ how do you get $n-2i=3$ $\endgroup$ Jul 6, 2019 at 21:23
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Suppose I of the 27 luggage are insured. The number of ways one of the 4 luggage is insured is $${4\choose1}I(27-I)(26-I)(25-I)$$
The number of ways none is insured is $$(27-I)(26-I)(25-I)(24-I)$$ The former is twice the latter. $$4I(27-I)(26-I)(25-I)=2(27-I)(26-I)(25-I)(24-I)$$

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