I am trying to figure out the math to get to an answer that's given to me right away, which is $1/3$. The question is asking what the probability of $B$ is with $P(A)=0.4$ and $P(A∪B)=0.6$, with $A,B$ being independent events.

I can't seem to come to one-third? Can someone explain it to me like I'm a three year old, since I have been on this for about a few hours?

Thanks, and much appreciated! Below is the setup I keep using:


which, when substituting in the given values and letting $P(B) = x$, yields

$$0.6 = 0.4 + x - P(A \cap B)$$

But I don't know what $P(A \cap B)$ is, I guess $0.4x$?

Eevee Trainer
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  • You have written the correct equation above (your guess is correct). Solving this equation should give you $x$. – Peter Foreman Mar 04 '19 at 23:28
  • Btw thanks. I don't know how to do most of the stuff you edited @Eevee Trainer – Smitty Mar 04 '19 at 23:41
  • It's a sort of math-rendering language called MathJax, which has a lot of derivatives from LaTeX - it's sort of how people can write the fancier math text you might see in a book, to put it in layman's terms. A quick reference on some of its syntax and such is here -- https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Eevee Trainer Mar 04 '19 at 23:42
  • @EeveeTrainer Okay, I just pulled it up, and bookmarked it. Thank you very much! – Smitty Mar 04 '19 at 23:44

5 Answers5


We know that, if $A,B$ are independent, then

$$P(A \cap B) = P(A)P(B)$$

We also know, through inclusion-exclusion,

$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

Combining the two, if we use inclusion-exclusion for $A,B$ independent,

$$P(A \cup B) = P(A) + P(B) - P(A)P(B)$$

In your problem, you are given $P(A \cup B)$ and $P(A)$. We note by factoring out the $P(B)$ in the previous expression,

$$P(A \cup B) = P(A) + P(B)(1 - P(A))$$

You should be able to take it from here, now it's just algebra.

Edit: (the remainder of the solution to address confusion that came up in later comments on other answers)

So, we solve for $P(B)$ by subtracting $P(A)$ from both sides, then dividing by $1 - P(A)$, thus getting

$$P(B) = \frac{P(A\cup B) - P(A)}{1 - P(A)}$$

We plug in the given values - $P(A) = 0.4, P(A \cup B) = 0.6$:

$$P(B) = \frac{0.6 - 0.4}{1 - 0.4} = \frac{0.2}{0.6} = \frac 1 3$$

Eevee Trainer
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Yes, if $A$ and $B$ are independent, then $\mathsf P(A\cap B)=\mathsf P(A)\,\mathsf P(B)$ .

Hence the formula you seek is $\mathsf P(A\cup B)=\mathsf P(A)+\mathsf P(B)-\mathsf P(A)~\mathsf P(B)$ rearranged to: $$\mathsf P(B)=\dfrac{\mathsf P(A\cup B)-\mathsf P(A)}{1-\mathsf P(A)}$$

Graham Kemp
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By definition if $A, B$ are independent events then $P(A \cap B) = P(A) P(B)$. Substituting in the inclusion-exclusion formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ you wrote gives $$P(A \cup B) = P(A) + P(B) - P(A) P(B) .$$ Now, you're given $P(A \cup B) = \frac{3}{5}$ and $P(A) = \frac{2}{5}$ substituting in the above equation gives an equation in $P(B)$ alone.

Travis Willse
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If P(A) and P(B) are independent then P(A)P(B)=P(AnB)


We know P(A)P(B)=0.4(0.2+P(AnB))=P(AnB)

Hence P(AnB)=2/15 and so P(B)=P(B\A)+P(AnB)=1/3

Joe Blogs
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Okay, I just found out what is wrong with what I am trying to do. Remember that -.4x? Combine that with x. Then, subtract the .4 from .6, and divide. So something like this, if I do the proper bold-ing and italics:

.6 = .4 + x - .4x

.6 = .4 + .6x

.2 = .6x

.33 = x

There, that should make more sense. Thank you guys for the help! That was actually really helpful, now I know something more from all of your help, so have a great day!

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