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Suppose that $P(A_1 \mid B) = 0.7$, $P(A_2 \mid B) = 0.4$, and $P(A_1 \cap A_2 \mid B) = 0.3$.

Given that B has occurred, find the probability that:
(a) at least one of the events $A_1$, $A_2$ occurs
(b) exactly one of the events $A_1$, $A_2$ occurs
(c) only $A_1$ occurs

so for a) $$P(A_1 \mid B) = \dfrac{P(A_1 \cap B)}{P(B)} = 0.7$$ $$P(A_2 \mid B) = \dfrac{P(A_2 \cap B)}{P(B)} = 0.4$$ $$P(A_1 \cap A_2 \mid B) = \dfrac{P(A_1 \cap A_2 \cap B)}{P(B)} = 0.3$$

We must find $$P(A1 \cup A2 \mid B) = \frac{P(A_1 \cup A_2 \cap B)}{P(B)}$$

I'm not sure how to find this probability.

b) We have to find $P(A_1 \cap A_2'\mid B) \cup P(A_1' \cap A_2 \mid B)$

Not really sure how to approach c). Is it as simple as $P(A_1 \mid B)$?

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  • $\begingroup$ Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site. $\endgroup$ Jan 23 '16 at 22:26
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You do not need to apply the formula of conditional probability, but only to use that $P(\cdot \mid B)$ is a probability measure and works fine as does $P(\cdot)$. So, for example the well known rule $$P(A_1\cup A_2)=P(A_1)+P(A_2)-P(A_1\cap A_2)$$ still holds when $B$ occurred (occurrence of $B$ does not change the rules of probability) and simply becomes $$P(A_1\cup A_2\mid B)=P(A_1 \mid B)+P(A_2\mid B)-P(A_1\cap A_2 \mid B)$$ Similarly for all other rules that you know. So,

  • (a) $P(A_1\cup A_2\mid B)=P(A_1 \mid B)+P(A_2\mid B)-P(A_1\cap A_2 \mid B)=0.7+0.4-0.3=0.8$
  • (b) $P(A_1\cup A_2 \mid B)-P(A_1 \cap A_2 \mid B)=0.8-0.3=0.5$
  • (c) $P(A_1 \cup A_2 \mid B)-P(A_2)=0.8-0.4=0.4$
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  • $\begingroup$ i had a feeling this was true but was trying to see if i could find a therom on it. thanks for the help ! $\endgroup$
    – Auron G
    Jan 23 '16 at 22:30
  • $\begingroup$ @AuronG You can prove by yourself that it is a probability measure under the assumption that $P(B)>0$ (which is not implied by the fact that $B$ occurred!!) $\endgroup$
    – Jimmy R.
    Jan 23 '16 at 22:34
  • $\begingroup$ @AuronG Also: $\frac{P((A_1\cup A_1)\cap B)}{P(B)}=\frac{P((A_1\cap B)\cup( A_1\cap B))}{P(B)}=\frac{P(A_1\cap B)+P(A_2\cap B)-P(A_1\cap A_2\cap B)}{P(B)}$ $\endgroup$ Jan 23 '16 at 23:17

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