MHB Probability of 2 Green Marbles

[mathdad]

Gilligan has 16 green marbles, 50 blue marbles and 60 red marbles. What is the probability of randomly selecting 2 green marbles?

What is the set up?

 

[Greg]

With or without replacement?

 

[mathdad]

With or without replacement?

Can you show me with both?

Someone suggested the following set up:

16C2 ÷ 136

Is this right?

 

[Greg]

Can you show me with both?

Well, what is the probability that the first marble selected will be green? With replacement, that's also the probability that the second marble selected will be green. Then what?

Without replacement, what do we have to do to the number of marbles when we determine the probability of the second marble being green?

Someone suggested the following set up:

16C2 ÷ 136

Is this right?

I don't see where '136' comes from. Shouldn't that be 126?

 

[Jameson]

Can you show me with both?

Someone suggested the following set up:

16C2 ÷ 136

Is this right?

Well often we see "combination" notation come up but it's important to think these problems through and not just use that.

With probability type questions, we usually have this situations: $$\frac{\text{# of desired outcomes}}{\text{# of possible outcomes}}$$. Let's start with the top half, the numerator.

1) Without replacement: For the first marble, how many choices do we have. For the second marble how many?

2) With replacement: same questions.

 

[HallsofIvy]

Gilligan has 16 green marbles, 50 blue marbles and 60 red marbles. What is the probability of randomly selecting 2 green marbles?

What is the set up?

With replacement: there are a total of 16+ 50+ 60= 126 marbles, 16 of them green. The probability the first marble drawn is green is 16/126= 8/63. That marble is replaced so on the second draw there are still 126 marbles, 16 of them green. The probability the second marble drawn is green is also 8/63. The probability both marbles are green is (8/63)(8/63)= (8/63)^2 which is about 0.0161.

Without replacement: there are a total of 16+ 50+ 60= 126 marbles, 16 of them green. The probability the first marble drawn is green is 16/126= 8/63. That marble is not replaced so now there are 125 marbles, 15 of them green. The probability the second marble drawn is green is 15/125= 3/25. The probability both marbles are green is (8/63)(3/25) which is about 0.0152.

 

[Jameson]

Can you show me with both?

Someone suggested the following set up:

16C2 ÷ 136

Is this right?

Well my questions still stand and I'm curious how you came up with this RTCNTC? :)

There is a way to write the solution in terms of combinations, but it's important to understand the concept. I'll give a hint: the total number of ways to choose ANY two marbles is 126C2, or $$\binom{126}{2}$$. That would be the denominator in my previous post. What about the numerator?

 

[mathdad]

Thank you everyone.