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I know there are similar questions out there, but I really need help understanding the problem I have.

A vase contains 100 glass balls. 20 are red, 10 are green, and 70 are neither green or red. I select two balls from the vase without replacement.

  • Let $R_1$ be the event that the first of the two selected is Red
  • Let $R_2$ be the event that the second of the two selected is Red
  • Let $G_1$ be the event that the first of the two selected is Green
  • Let $G_2$ be the event that the second of the two selected is Green
  • Let A be the event that at least one of the two selected is Red
  • Let B be the event that at least one of the two selected is Green

Question: P(A$\cap$B) = ?

MY ATTEMPT:

Originally, my answer was P(A$\cap$B) = P(A)*P(B) = 0.076

This seems to be wrong since they are not independent (not sure why, but I read this in a similar question).

I tried looking at the problem from a different perspective, e.g. looking at the inverse of P(A) and P(B), but I couldn't see how to make it work.

Lastly, since P(A$\cap$B) = P(A)*P(B|A), I tried to reason out the probability for P(B|A), but again my intelligence had forsaken me.

How would I do this problem? Please explain in the simplest way possible, I am only in an introductory statistics course and haven't covered much. To be specific, I'm reading Elementary Statistics (10th Ed.) by Mario F. Triola and my class has only made it to chapter 4 (when to add probabilities, when to multiply probabilities, conditional probabilities).

If someone could work out the problem and explain what's going on, it will help me tremendously with the rest of my assignment, since the next problems are dependent on this value. Unfortunately, due to the teaching style of my professor, I must learn on my own and I am at a loss right now.

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    $A\cap B=(R_1\cap G_2)\cup(G_1\cap R_2)$ and the events $R_1\cap G_2$ and $G_1\cap R_2$ are mutually exclusive. So if you can find the probabilities of them then... Any idea's of your own now? – 2017-02-12
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    That's all I had to do? I was overthinking this the entire time? – 2017-02-12
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    Yes. Since there are only $2$ balls you can leave out the words "at least" in your question. Can you find the probabilities of the events I mentioned? – 2017-02-12
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    Yes, everything is smooth sailing again. I had this equation written down at some point but thought it was incorrect. Thank you! – 2017-02-12

2 Answers 2

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This seems to be wrong since they are not independent (not sure why, but I read this in a similar question).

If you pick a red ball out of the urn, there are fewer red balls in the urn so the probability of then picking a green ball is slightly increased.   Hence there is a dependency.

Lastly, since $P(A\cap B) = P(A)\cdot P(B\mid A)$, I tried to reason out the probability for $P(B\mid A)$, but again my intelligence had forsaken me.

Oft times the joint probability may be easier to find than either conditional.   Observe that you are after the probability for selecting one from the twenty red and one from the ten green when selecting any two from all hundred balls.

$$\mathsf P(A\cap B)~=~\dfrac{\dbinom {20}1\dbinom{10}1}{\dbinom{100}2}~=~\dfrac{2\cdot 20\cdot 10}{\quad100\cdot 99}$$

Which then allows you to find the conditional probability of selecting at least one green when given having selected at least one red.

$$\mathsf P(B\mid A) = \dfrac{\dbinom {20}1\dbinom{10}1}{\dbinom {20}1\dbinom{10}1+\dbinom {20}2\dbinom{10}0}~=~\dfrac{20}{39}$$

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    The total number of balls is $100$ rather than $70$. – 2017-02-12
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    If P(A) = 0.36 and P(A$\cap$B) = .04, why is P(B|A) = ~0.51 and not my obtained value from the equation, ~0.111? – 2017-02-12
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    Because $\mathsf P(A) ~=~ 13/165 ~\approx~ 0.079$ – 2017-02-13
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Here: $$A\cap B=(R_1\cap G_2)\cup(G_1\cap R_2)$$ The events $R_1\cap G_2$ and $G_1\cap R_2$ are mutually exlclusive so that $$\Pr(A\cap B)=\Pr(R_1\cap G_2)+\Pr(G_1\cap R_2)$$ We find $$\Pr(R_1\cap G_2)=\Pr(R_1)\Pr(G_2\mid R_1)=\frac{20}{100}\frac{10}{99}$$ and $$\Pr(G_1\cap R_2)=\Pr(G_1)\Pr(R_2\mid G_1)=\frac{10}{100}\frac{20}{99}$$ So we end up with:$$\Pr(A\cap B)=2\times\frac{10}{100}\frac{20}{99}$$

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    Are they mutually exclusive because you can't pick red and green at the same time? – 2017-02-12
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    They are mutually exclusive because it cannot happen that the first is red and the second green ($R_1\cap G_2$) and also the first is green and the second red ($G_1\cap R_2$). – 2017-02-13