0
$\begingroup$

Out of 15 phones, 5 are cordless, 5 are corded and 5 are cellular. Each component is randomly selected and allocated numbers 1, 2..., 15 to establish the order in which they are serviced.

a) What is the probability that all the cordless phones are among the first ten to be serviced?

b) What is the prob. that after servicing 10 of these phones, phones of only 2 of the three types remain to be serviced?

For a) I did [(5 choose 5)*(10 choose 5)] / (15 choose 10). I was wondering if using combinations was the correct approach instead of permutations.

For b, I was confused on how to calculate this. I initially did: [(5 choose 0)*(10 choose 5)] / (15 choose 5). However, I don't think this is correct. Do I need to use (3 choose 2) since I'm considering which types of phones are left to be serviced?

Thanks for any help!

2 Answers 2

1

a) You want to count the probability for selecting $5$ from $10$ places for cordless phones the when selecting $5$ from $15$.

[edit: You have the probability for selecting all of the cordless phones and five from the ten others when selecting any ten of the fifteen. That is okay too. ]

b) You have calculated probability that all five cordless phones have been serviced, which is also the probability that all five of each of the other two types have been serviced. However, if you just added you would be over counting common cases; so you must exclude the probability for selecting all five of two types at once.

(IE use the Principle of Inclusion and Exclusion)

  • 0
    I couldn't quite understand "selecting 5 from 15" part of **a**. – 2017-02-05
  • 0
    @trueblueanil There are *fifteen* positions for the phones to be serviced, then select, without bias, *five* from those *fifteen* positions for the cordless phones. We want the probability that those *five* positions are selected from first *ten*. – 2017-02-06
  • 0
    Oh, I see. You focused your attention entirely on the cordless phones. The simpler, the better ! (+1) – 2017-02-06
0

a

OK.

b

You can avoid PIE by realizing that there are just $2$ valid patterns,
$5-4-1$ and $5-3-2,$ each with $3!$ permutations,

thus $Pr = \dfrac{3!\left[\dbinom55\dbinom54\dbinom51 + \dbinom55\dbinom53\dbinom52 \right]}{\dbinom{15}{10}}$