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I have a question that I am completely clueless about. Please help me understand.

$P(a)= 0.20$ $P(b)=0.30$ and $P( a\mbox{ or }b)=0.48$. Don't assume anything about the events.

  1. Find $P( a\mbox{ and }b)$
  2. Find $P ( a\mbox{ and }b')$
  3. Find $P (\mbox{neither }a\mbox{ nor }b)$
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2 Answers 2

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$1.$ Note that $P(a \cup b)=P(a)+P(b)-P(a \cap b)$ or in your terms, P(a or b)=P(a)+P(b)-P(a and b).

Thus, answer of Q.1 is $P(a \cap b)=0.02$.

$2.$ Note that P ( a and b')$=P(a\cap b')=P(a)-P(a\cap b)=0.2-0.02=0.18$

$3.$ Note that P (neither a nor b)$=1-P(a \cup b)=1-0.48=0.52$

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Hints:

$1.$ You have probably been introduced to the formula $$\Pr(A\cup B)=\Pr(A)+\Pr(B)-\Pr(A\cap B).\tag{1}$$

In Problem $1$, you are given two of the three probabilities mentioned in Formula (1).

$2.$ In general, we have $$\Pr(X)=\Pr(X\cap Y)+\Pr(X\cap Y').$$
A Venn diagram makes this reasonably clear.

$3.$ The event "neither $A$ nor $B$" is the event $(A\cup B)'$.

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