Given that P(A and B)=0.1 and P(A and B')=0.4 find P(A or B') if A and B are independent.
The ans is 0.9
Please tell me the rule you use in the problem.
$\endgroup$ 23 Answers
$\begingroup$Two events $X$ and $Y$ are independent if $$ \Bbb P(X\text{ and }Y) = \Bbb P(X)\cdot\Bbb P(Y) $$ We also have the general formulas \begin{align*} \Bbb P(X\text{ or }Y) &= \Bbb P(X)+\Bbb P(Y)-\Bbb P(X\text{ and }Y) \\ \Bbb P(X) &= 1-\Bbb P(X^\prime) \end{align*} Can you use these formulas to solve your problem?
$\endgroup$ 2 $\begingroup$I already solved the problem. NVM
0.1 = P(A)P(B)
P(A)=0.4+0.1
0.1 = 0.5P(B)
P(B) = 0.2
P(B')=0.8
P(A or B')=P(A)+P(B')-P(A and B')
= 0.5+0.8-0.4
=0.9
$\endgroup$ $\begingroup$By simply using the Venn diagram. It is clear that
P(A or B') = P(A) + P(B') -P(A and B') and P(A and B') = P(A) - P(A and B), this implies that 0.4 = P(A) - 0.4, Therefore P(A) = 0.5
A and B are independent events, this means that P(A and B) = P(A)*P(B) Therefore 0.1 = 0.5*P(B), P(B) = 0.2 Consequently, P(B') = 1 - 0.2 = 0.8
Finally, P(A or B') = 0.5 + 0.8 - 0.4 = 0.9
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