1. ## Probability Question Check

Hi,

15 phones have just been received at an authorized service center. 5 are cellular, 5 are cordless, and 5 are corded. Suppose service tickets have been issued for these phones allocating all numbers 1 through 15 indicating the order in which they will be serviced. Find the probabilities that:

after having repaired 10 phones, there are phones of exactly two of the three types remaining to be serviced.

I computed the answer to be [3 x [(10 C 5) - 2]]/(15 C 10)

Is this correct

2. Originally Posted by shogunhd
Hi,

15 phones have just been received at an authorized service center. 5 are cellular, 5 are cordless, and 5 are corded. Suppose service tickets have been issued for these phones allocating all numbers 1 through 15 indicating the order in which they will be serviced. Find the probabilities that:

after having repaired 10 phones, there are phones of exactly two of the three types remaining to be serviced.

I computed the answer to be [3 x [(10 C 5) - 2]]/(15 C 10)

Is this correct
I get the same answer that you do

(but used a different approach. My numerator is:

$\displaystyle 3\left[ {5 \choose 4} \cdot {5 \choose 1} + {5 \choose 3} \cdot {5 \choose 2} + {5 \choose 2} \cdot {5 \choose 3} + {5 \choose 1} \cdot {5 \choose 4}\right]$

$\displaystyle = 3\left[ 2 \left\{ {5 \choose 4} \cdot {5 \choose 1} + {5 \choose 3} \cdot {5 \choose 2} \right \} \right]$

$\displaystyle = 6 \left \{ {5 \choose 4} \cdot {5 \choose 1} + {5 \choose 3} \cdot {5 \choose 2} \right \} = 6 (5^2 + 10^2) = 750$ )