# Thread: Combination Problems

1. ## Combination Problems

there are 12 resistors. 3 are defective. if you choose 4 what is the probability that you'll get:

no defective resistors
1 defective resistor
3 defective resistors

15 insurance policies: 8 life, 5 auto, 2 home. if you choose 3, what is the probability that you'll get
all life policies
both home policies
1 of each
all auto policies
2 life and 1 auto policy

Help please.

2. Hello, Nellie!

There are 12 resistors; 3 are defective.
If you choose 4 what is the probability that you'll get:

(a) no defective resistors . (b) 1 defective resistor . (3) 3 defective resistors
There are: . ${12\choose4} = 495$ possible samples.

(a) "no defective" means all four are good.
There are: . ${9\choose4} = 126$ ways to choose four good resistors.
Therefore: . $P(\text{0 d{e}f}) \:=\:\frac{126}{495} \;=\;\frac{14}{55}$

(b) "one defective" means 1 defective and 3 good resistors.
There are: . ${3\choose1}{9\choose3} \:=\;3\cdot84 \:=\:252$ ways.
Therefore: . $P(\text{1 d{e}f}) \:=\:\frac{252}{495} \;=\;\frac{28}{55}$

(c) "3 defective" means all 3 defective resistors and 1 good resistor.
There are: . ${3\choose3}{9\choose1} \:=\:1\cdot9 \:=\:9$ ways.
Therefore: . $P(\text{3 d{e}f}) \;=\;\frac{9}
{495} \;=\;\frac{1}{55}$

3. ## Soroban

Thank you! and you explained it so clearly!