Probability of Independent Event

TheJason

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#1
Probability of Single Choice 2 in Group 2 = (Amount of Choice 1/Amount of Group 1)(Amount of Choice 1/Amount of Group 2) + (Amount of Choice 2/Group 1)(Amount of Choice 2/Amount of Group 2)

@MarkFL Well, we are assuming two groups and two choices, but what if there were more groups and/or choices?

Example problem:

One bag has 2 orange balls and 1 grey ball. Another bag has 10 orange balls and 6 grey balls. What is the chance after one ball is moved from one to the other - that one picked ball in the 2nd group is grey?

Choice 1 is orange and Choice 2 is grey.

Probability of single grey ball picked out of bag 2 = (2/3)(11/17) + (1/3)(7/17)

:) - Notice how in the numbers pertaining to the 2nd bag, the amount of choice and the amount of group are increased by 1?
 
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MarkFL

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#2
Okay, so you get:

\(\displaystyle P(X)=\frac{29}{51}\)

If I were to reason my way though this one, I would first look at the case where the ball moved from the first bag is orange:

\(\displaystyle P(A)=\frac{2}{3}\cdot\frac{6}{17}=\frac{4}{17}\)

Then the case where the ball moved from the first bag is grey:

\(\displaystyle P(B)=\frac{1}{3}\cdot\frac{7}{17}=\frac{7}{51}\)

And so, we have:

\(\displaystyle P(X)=P(A\text{ or }B)=P(A)+P(B)=\frac{19}{51}\)

Where did you get 11/17 from?
 
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TheJason

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Where did you get 11/17 from?
Bag 2 is increased overall by 1 and the orange amount in the bag 2 is increased by 1.

Note, this stuff is based on another problem I saw in a textbook.
 
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