# Probability of Independent Event

#### TheJason

Staff member
Moderator
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

##### La Villa Strangiato
Staff member
Moderator
Math Helper
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?

anemone