Finding $P(B)$ if $A,B$ are independent and given values for $P(A),P(A\cup B)$

Question

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:

$$P(A∪B)=P(A)+P(B)-P(A∩B)$$

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 $0.4x$?

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 — PrincessEev, 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

PrincessEev · Accepted Answer · 2019-03-05T00:28:13.307

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 - $P(A) = 0.4, P(A \cup B) = 0.6$:

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

Okay, that makes some sense now. Thank you! Really helps me out for sure. — Smitty, Mar 04 '19 at 23:29

score 0 · Answer 2 · answered Mar 04 '19 at 23:26

0

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: $$\mathsf P(B)=\dfrac{\mathsf P(A\cup B)-\mathsf P(A)}{1-\mathsf P(A)}$$

answered Mar 04 '19 at 23:26

Graham Kemp

129,094

2

The OP already wrote this formula in their working – Peter Foreman Mar 04 '19 at 23:29
1

And, Yes, that is the formula they seek. – Graham Kemp Mar 04 '19 at 23:30

score 0 · Answer 3 · answered Mar 04 '19 at 23:28

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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.

answered Mar 04 '19 at 23:28

Travis Willse

99,363

So do you suggest that P(A_U_B) is zero? – Smitty Mar 04 '19 at 23:38
Why would $P(A \cup B)$ be zero? After all, it's given in the problem as $0.6.$ – PrincessEev Mar 04 '19 at 23:40
@Smitty No, I don't suggest that---like Eevee says, it's given to be $\frac{3}{5}$. What equation do you get when you substitute the two given values in the equation in my answer? – Travis Willse Mar 04 '19 at 23:46
@Smitty You can typeset $P(A \cup B)$ with the LaTeX code P(A \cup B). – Travis Willse Mar 04 '19 at 23:47
@EeveeTrainer meant the other symbol, I can't seem to type properly today. – Smitty Mar 04 '19 at 23:48
@Travis Okay, and thank you as well! – Smitty Mar 04 '19 at 23:50
@Smitty No, as I say in my answer, independence of $A, B$ says exactly that $P(A \cap B) = P(A) P(B)$. – Travis Willse Mar 04 '19 at 23:50
(And you're welcome!) – Travis Willse Mar 04 '19 at 23:51
Okay, I'll try it right now, hopefully it comes out right! – Smitty Mar 04 '19 at 23:52
Oh I just realized, did I mention what the answer was? Edit: I did, but nothing comes up as 1/3.. – Smitty Mar 04 '19 at 23:53
@Smitty I made an addendum to my original answer completing the algebra in case you want to see my work. I'm not sure what you're doing, but I am getting $1/3$ as intended. – PrincessEev Mar 05 '19 at 00:29
@Smitty Substituting gives $\frac{3}{5} = \frac{2}{5} + P(B) - \frac{2}{5} P(B)$, and rearranging gives $\frac{1}{5} = \frac{3}{5} P(B)$, so $P(B) = \frac{1}{3}$. – Travis Willse Mar 05 '19 at 01:00

score 0 · Answer 4 · answered Mar 05 '19 at 00:14

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

P(B\A)=P(AUB)-P(A)=0.2

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

answered Mar 05 '19 at 00:14

Joe Blogs

5

score 0 · Answer 5 · answered Mar 06 '19 at 00:14

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!

Finding $P(B)$ if $A,B$ are independent and given values for $P(A),P(A\cup B)$

5 Answers5