Question 1194683: A sample of 5 is to be chosen from a batch of 6 resistors and 9 transistors. If the selection is made randomly, what is the probability that the sample consists of 3 resistors and 2 transistors?
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
There are n = 6+9 = 15 items total
We want to select r = 5 of them, in which order does not matter.
Use the nCr combination formula.
n = 15
r = 5
n C r = (n!)/(r!(n-r)!)
15 C 5 = (15!)/(5!*(15-5)!)
15 C 5 = (15!)/(5!*10!)
15 C 5 = (15*14*13*12*11*10!)/(5!*10!)
15 C 5 = (15*14*13*12*11)/(5!)
15 C 5 = (15*14*13*12*11)/(5*4*3*2*1)
15 C 5 = (360360)/(120)
15 C 5 = 3003
There are 3003 ways to form a sample of 5 items, where order doesn't matter.
This sample of five may or may not fit the criteria of "3 resistors and 2 transistors".
Let's focus on just the resistors.
We have n = 6 resistors and want to select r = 3 of them.
Use the nCr formula. I'll skip showing the steps, but you should show the steps to your teacher. Follow the template shown above.
You should find the following
6 C 3 = 20
There are 20 ways to select the three resistors from a pool of six total.
You should also find that
9 C 2 = 36
meaning there are 36 ways to pick the two transistors
Once again, I'm skipping steps, but you should show them to your teacher.
We have...- 20 ways to pick the three resistors (from a pool of six)
- 36 ways to pick the two transistors (from a pool of nine)
So,
There are 20*36 = 720 ways to pick the three resisters and two transistors in any order.
We have A = 720 ways of getting what we want (picking the three resisters and two transistors) out of B = 3003 ways to get a sample of five items.
Divide those values to get the probability we want.
A/B = 720/3003 = (240*3)/(1001*3) = 240/1001
Answer: 240/1001
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