Coins in a bag

C

colerelm

New member
Joined
Oct 24, 2011
Messages
33
There are 8 coins in a box. 5 are red, 2 are blue, and 1 is green.
Without watching, you randomly grab 3 coins from the box. What is
the chance that at you end up with at least 1 red coin and at least 1 blue
coin ?


Can anyone guide me in how I would do this problem?
 
What's the probability of drawing a red on the first draw?
What's the probability of drawing a blue on the first draw?
What's the probability of drawing a green on the first draw?
 
Hello, colerelm!

There is no formula for this problem.
We must make a List.

There are 8 coins in a box. 5 are red, 2 are blue, and 1 is green.
Without looking, you randomly grab 3 coins from the box.
What is the probability that at you get at least 1 red coin and at least 1 blue coin ?
Click to expand...

There are: .\(\displaystyle {8\choose3} \,=\,56\) possible outcomes.


There are three successful outcomes:

[1] one Red, one Blue, one Green.
. . .There are: .\(\displaystyle {5\choose1}{2\choose1}{1\choose1} \:=\:5\cdot2\cdot1 \:=\:10\) ways.

[2] two Red, one Blue.
. . .There are: .\(\displaystyle {5\choose2}{2\choose1} \:=\:10\cdot2 \:=\:20\) ways.

[3] one Red, two Blue.
. . .There are: .\(\displaystyle {5\choose1}{2\choose2} \:=\:5\cdot1 \:=\:5\) ways.

Hence, there are: .\(\displaystyle 10+20+5 \:=\:35\) successful outcomes.


Therefore: .\(\displaystyle P(\,\ge \text{1 Red }\text{ and }\ge\text{1 Blue}) \:=\:\dfrac{35}{56} \:=\:\dfrac{5}{8}\)
 
So what about the possibility of one green + two red or one green + two blue?

edit: nevermind, I forgot that isn't what the question was asking. Thanks!
 
You must log in or register to reply here.
Top