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When the health department tested private wells in a county for two impurities commonly found in drinking water, it found that 20% of the wells had neither impurity, 40% had impurity A, and 50% had impurity B. (Obviously, some had both impurities.) If a well is randomly chosen from those in the county, find the probability distribution for Y , the number of impurities found in the well.

The answers:

P($Y=0$) = 0.2

P($Y=1$) = P($A$) + P($B$) - P($A\cap B$) ** , which is 0.7

P($Y=2$) = P($A\cap B$), which is 0.1

However, my friends and I disagreed about how to find P($A\cap B$). If you're given P(A), which is 0.4, and you're given P(B), which is 0.5, then in order to find P($A\cup B$), don't you just do $0.4 + 0.5 - 0.4(0.5$) as according to the theorem?

But then the answer when you multiply the two is 0.2, which is NOT 0.1 as given.

My friend, on the other hand, subtracted 0.1 from 0.4 and from 0.5, so she added 0.3 + 0.4 together in order to get the result 0.7. But if you're using the theorem, you can't subtract from both. What are we doing wrong? Did the 0.2 have something to do with this?

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$P(A\cap B) \neq P(A)P(B)$ in general. You cannot use the multiplication rule $P(A\cap B) = P(A)P(B)$ unless you are told that, or you can make a persuasive argument to the effect that, the events $A$ and $B$ are independent. – Dilip Sarwate Sep 26 '12 at 19:09
Your equation $P\{Y = 1\} = P(A) + P(B) - P(A\cap B)$ is incorrect. $\{Y = 1\}$ is the event that exactly one of $A$ and $B$ occurred, and this event $A\oplus B$ has probability $$P(A\oplus B) = P(A)+P(B)-\mathbf{2}P(A\cap B)$$ Your problem statement allows you to compute $$P(A\cup B) = P\{\text{at least one of}~A~\text{and}~B\}=1-P(A^c\cap B^c)-1-0.2=0.8.$$ But, $P(A\cup B)=0.8=P(A)+P(B)-P(A\cap B)=0.4+0.5-P(A\cap B)$ giving $P(A\cap B)=0.1$ and hence $$P(A\oplus B)=P(A\cup B)-P(A\cap B)= 0.8-0.1=0.7.$$ – Dilip Sarwate Sep 26 '12 at 20:09
up vote 4 down vote accepted

The main problem is that $P(A\cap B)=P(A)P(B)$ is true only when $A$ and $B$ are independent events, and according to the statement of the problem $P(A\cap B)=0.1\neq 0.2=0.4\cdot 0.5=P(A)P(B)$, i.e. it is implicity saying that both events are not independent.

In conclusion, $P(A\cap B)=P(A)P(B)$ is not always true -only when the considered events are independent-. And thus, for calculating $P(A\cup B)$ you shouls apply that $$0.2=P(\overline{A}\cap \overline{B})=P(\overline{A\cup B})=1-P(A\cup B)$$ which gives that $P(Y\geq 1)=P(A\cup B)=0.8$. And here, the rest is straighforward.

However, note that $Y=1$ is equivalent to $(\overline{A}\cap B)\cup (A\cap \overline{B})$ not to $A\cup B$, since this last means that impurity A is found and not B, that impurity B is found and not A or that both impurities are found.

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Since $P(Y=0)=.2$, then $P(A\cup B)=P(Y=1\text{ or }Y=2)=.8$. Hence, by the theorem, $$P(Y=2)=P(A\cap B)=P(A)+P(B)-P(A\cup B)=.1,$$ and so $$P(Y=1)=P(Y=1\text{ or }Y=2)-P(Y=2)=.7.$$ On the other hand, we can say that $$P(Y=1)=P(A\cap B^c)+P(B\cap A^c)=\bigl(P(A)-P(A\cap B)\bigr)+\bigl(P(B)-P(A\cap B)\bigr)=(.4-.1)+(.5-.1)=.7.$$ I think the source of your confusion about your friend's approach lies in the fact that while (in general) $P(A\cup B)\neq P(A)+P(B)-2P(A\cap B)$, your friend was trying to calculate $P(Y=1)$--which is the probability that a well has exactly one impurity--and not $P(A\cup B)$--which is the probability that a well has one or both impurities.

You were attempting to use $P(A\cap B)=P(A)P(B)$. We can only do that if A and B are independent--in other words, if $50\%$ of the wells with impurity A also had impurity B, and $40\%$ of the wells with impurity B also had impurity A. We don't know this, however. Perhaps the impurities correspond to two different environmental factors, and so it would be less likely for a well with one impurity to have the other as well. Or perhaps one of the impurities can actually cause the other in the right circumstances, so wells with the first impurity are actually more likely to have the second than they would otherwise be. We simply don't know, so we can't say that $P(A\cap B)=P(A)P(B)$.

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$$ a = P(A,\bar{B})\\ b = P(\bar{A},B)\\ c = P(A,B)\\ 0.2 = P(\bar{A},\bar{B}) $$

Now, $a+b+c=0.8$

Also, $a+c=0.4$

And, $b+c=0.5$

This gives: $b=0.4$, $a=.3$, and $c=0.1$

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