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Suppose A and B are events with P(A) 0.4 , P(B) 0.6 and P(A and B) 0.25 . Calculate the probability P(A complement union B).

A 0.25

B 0.65

C 0.75

D 0.85

What I tried?-

P(A union B)=P(A)+p(B)-P(A and B) i.e=0.4+0.6-0.25=0.75. I am stuck after this. i know this is simple but I am unable to find the right approach. enter image description here

Below is the diagram that I created after solving up to here.

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    $\begingroup$ Start by drawing a little diagram with all the cases (an array with A and not A as lines, B and not B as columns), you'll see how it works. $\endgroup$ – Joce Nov 2 '15 at 16:30
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Your graph is fine. Now you can mark the area of $\overline A$. I did this with red lines. It is just the whole rectangle without $A$.

enter image description here

After that you mark the area of $B$

I did this with blue lines.

enter image description here

Now the union of this two areas are the whole rectangle without the unmarked area. Because of the values you already have inserted it can be easily seen (calculated) what $P(\overline A \cup B)$ is.

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  • $\begingroup$ thanks for such a fine explanation. sorryy I am unable to figure out how to do calculation after this. $\endgroup$ – RajSharma Nov 3 '15 at 0:42
  • $\begingroup$ The unmarked area is $0.15$. And the area of the rectangle is $1$. Therefore the marked area is $1-0.15$. $\endgroup$ – callculus Nov 3 '15 at 0:49
  • $\begingroup$ Thanks it seems. P(A complement) is equal to P(A comp Union B) in this case. $\endgroup$ – RajSharma Nov 3 '15 at 0:51
  • $\begingroup$ Not really. $P(\overline A)=1-P(A)=1-0.4=0.6$ and $P(\overline A \cup B)=1-0.15=0.85$ It is comprehensible ? In your exercise the area of $A \cap B$ (blue marked only) is included, which is $0.25$. $\endgroup$ – callculus Nov 3 '15 at 0:55
  • $\begingroup$ thanks well understood now. Thanks a lot. $\endgroup$ – RajSharma Nov 3 '15 at 1:00

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