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The information I have is: $$P (A) = 3/8$$ $$P (B) = 1/2$$ $$P (A\cap B) = 1/4$$ And I have to calculate:

  1. $P ( A^c )=5/8$
  2. $P ( B^c )=1/2$
  3. $P ( A \cup B )=5/8$
  4. $P ( A^c\cap B^c )=3/8$
  5. $P ( A^c\cap B )=1/8?$

So, my question is about $(5)$. It is ok to say that If $P (A)

My calculation is $P ( A^c \cap B )=P(B-A)=P(B)-P(A)$. But $P (A)\neq P(A\cap B)$ , so it's mean that $A\nsubseteq B$?

How I can calculate it?

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    Nope: $A\subset B~\implies~\mathsf P(A)\leq \mathsf P(B)$, but the converse cannot be assured. – 2017-01-25
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    That is one of the answers I was looking for. Thank you. – 2017-01-25

3 Answers 3

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Use $$\mathbb{P}(B) = \mathbb{P}(B \cap A) + \mathbb{P}(B \cap A^c)\text{.}$$

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    Does not address question in title. – 2017-01-25
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Consider the following setup:

$$\begin{cases} \Omega = \{1,2\} \\ P(\{1\}) = 1/3 \\ P(\{2\}) = 2/3 \end{cases}$$

Let $A=\{1\}$ and $B=\{2\}$. Clearly there is no inclusion despite inequality of probability.

In fact, in general let $A$ be an event with $P(A)<1/2$ Then with $B=A^c$ we have

$$P(B)>1/2>P(A)$$

and clearly there is no inclusion.

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If $A\subseteq B$ then $A\cap B = A$ so you would have $P(A) = P(A\cap B)$ which is not true in your case.

If $P(A)