-1

Question: Fifteen telephones have just been received at an authorized service center. Five of these telephones are cellular, five are cordless, and the other five are corded phones. Suppose that these components are randomly allocated the numbers $1$, $2$, . . . , $15$ to establish the order in which they will be serviced. (Round your answers to four decimal places.) What is the probability that after servicing ten of these phones, phones of only two of the three types remain to be serviced?

Answer: $$\frac{3\cdot(^{5}\rm C_1\cdot ^{5}\rm C_4+^{5}\rm C_2\cdot^{5}\rm C_3+^{5}\rm C_3\cdot^{5}\rm C_2+^{5}\rm C_4\cdot^{5}\rm C_1)}{^{15}\rm C_{10}}$$

My question: Why isn't the denominator $^{15}\rm C_5$ when the numerator consists of certain groups of $5$ phones.

Cheese Cake
  • 1,143
IWantToLearn
  • 25
  • Because we are randomly choosing 10 phones out of 15 phones to be serviced. – insipidintegrator Sep 13 '22 at 06:44
  • First, the numerator does not consist of groups of phones, the numerator is just a number. Second, it doesn't matter if you write the denominator as ${15}C{10}$ or ${15}C_5$, because ${15}C_{10}=_{15}C_5$. – bof Sep 13 '22 at 06:59

2 Answers2

3

Answer: $$\frac{3\cdot(_{5}\rm C_1\cdot_{5}\rm C_4+_{5}\rm C_2\cdot_{5}\rm C_3+_{5}\rm C_3\cdot_{5}\rm C_2+_{5}\rm C_4\cdot_{5}\rm C_1)}{_{15}\rm C_{10}}$$ My question: Why isn't the denominator $_{15}\rm C_5$ when the numerator consists of certain groups of $5$ phones.

You're correct, but note that $$ {15\choose 5}={15!\over10!5!}={15\choose 10}. $$ You can think of it like this: we could just as well count all the ways that the first ten phones got all five of one type, ${5\choose5}=1$, and then some each of the other two types, which would give us exactly the same denominator and numerator, so the two are actually interchangeable.

Suzu Hirose
  • 11,660
0

The denominator is 15C10 because the question says what is the probability "after servicing ten of these phones", which implies it's 15C10 instead of 15C5.

Yongjian P.
  • 129