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In a bag there are $15$ balls.Five are green, five are yellow and five are white. The balls of each color are numbered from $1$ to $5$.

Now,suppose that in the bag there are only some of the $15$ balls. In these new conditions one ball is drawn from the bag, and we have:

-the probability of this ball be yellow is $50\%$

-the probability of this ball have the number $1$ is $25\%$

-the probability of this ball be yellow or have the number $1$ is $62,5\%$

Prove that the yellow ball with the number $1$ on its face is in the bag.

First I set two different events.

A-"get a yellow ball"

N-"get a ball with the number $1$ on its face"

Then I thought if I could prove that $P(A \cap N)>0$, I'll prove that there is a yellow ball with the number $1$ on its face.It's clear that $P(A)+P(N)\neq P(A \cup N)$ so A and N are not mutually exclusive and we know that $P(A \cup N)=P(A)+P(N)-P(A \cap N)$. By cumputing the expression with the correct values I achieved that $P(A \cap N)=12,5 \%$.So its prove that there is a yellow ball with the number $1$ on its face in the bag.

This is enough? Thanks again.

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I didn't know that was recomended to choose the best awswer.It's done. –  João Feb 27 '12 at 23:01
    
Yes, if there is an answer that answers your question, you should choose the best. Thanks for doing that! –  Graphth Feb 27 '12 at 23:21

1 Answer 1

up vote 0 down vote accepted

Yes, that looks right. Well done.

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