A bag contains 3 red, 4 white, 5 green balls...

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Kulla_9289

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A bag contains 3 red, 4 white, 5 green balls. Three balls are selected without replacement. Find the probability that the three balls chosen are:
a) all red
b) all green
c) one of each colour


My answers:
a) 1/40
b) 1/8
c) 13/60

Thanks in advance.
 
Kulla_9289 said:
A bag contains 3 red, 4 white, 5 green balls. Three balls are selected without replacement. Find the probability that the three balls chosen are:
a) all red
b) all green
c) one of each colour


My answers:
a) 1/40
b) 1/8
c) 13/60

Thanks in advance.
Click to expand...
Please show your work, so we can see what errors you are making.
 
In your tree, you appear to be selecting only two balls, not three; and your denominator drops from 12 to 10 after selecting the first.

1650724529456.png​

Can you explain your thinking there? I think you may have just omitted the second set of branches.

Also, you seem to be misinterpreting "one of each color"; what you are adding appears to be the probabilities of "all the same color".
 
Kulla_9289 said:
Yep, I understood and fixed that. How would you do "one of each color" then?
Click to expand...
You'd need to make a new tree. For the first pick, you would start out the same as the previous tree, but this time, the second and third picks are a different colour so that you have one of each colour from all 3 picks.
 
Kulla_9289 said:
Yep, I understood and fixed that. How would you do "one of each color" then?
Click to expand...
Do you know any methods other than making a tree? If you know about permutations, that could provide a quicker way.

But with a tree, you'll want to add up the probabilities of all outcomes that have one of each color. Show us your corrected tree if you need help with this.
 
I saw a medium in my further mathematics book, but we haven't studied that yet. Yeah, I did it, and I have gotten 3/11. Is that correct?
 
Kulla_9289 said:
I saw a medium in my further mathematics book, but we haven't studied that yet. Yeah, I did it, and I have gotten 3/11. Is that correct?
Click to expand...
Yes, the answer is 3/11.

Just for fun, I'll tell you how I got that number without a tree!

There are 6 different ways to order three colors. (This is called 3 factorial, or the permutations of 3 items, but you could just count: RWG, RGW, WRG, WGR, GRW, GWR.) For each order, you can choose the R in 3 ways, the W in 4 ways, and the G in 5 ways, for a total of 6*3*4*5 ways to choose one from each color. Divide this by the total number of ways to choose three of the 12 balls, which is 12*11*10 (this is the permutations of 12 taken 3 at a time), and you get the probability as
[math]\frac{6\cdot3\cdot4\cdot5}{12\cdot11\cdot10}=\frac{3}{11}[/math]
 
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