# Find probability that only one event will occur

I have a problem with such simple task:

Probabilities of two independent events $A_1$ and $A_2$ are equal repectively $p_1$ and $p_2$. Find the probability that only one of the events will occur.

The answer given in the book is $P=p_1+p_2-2 \cdot p_1 \cdot p_2$

However I get the different answer. The result from above is got when such way of solving is applied:

$$P((A_1 \cap A_2') \cup (A_1' \cap A_2))=P(A_1 \cap A_2') + P(A_1' \cap A_2)$$

and so on. I don't udnerstand why there is no substraction of common part of events, as there is not mentioned that those events are mutually exclusive. Shouldn't the first step be:

$$P((A_1 \cap A_2') \cup (A_1' \cap A_2))=P(A_1 \cap A_2') + P(A_1' \cap A_2)- P((A_1 \cap A_2') \cap (A_1' \cap A_2))$$

?

• Your first step is not wrong. Just take a closer look at the last term (cf. Brian's answer). – caligula Oct 31 '13 at 15:31

The events $A_1\cap A_2'$ and $A_1'\cap A_2$ are mutually exclusive:

$$(A_1\cap A_2')\cap(A_1'\cap A_2)=(A_1\cap A_1')\cap(A_2\cap A_2')=\varnothing\cap\varnothing=\varnothing\;$$

You can certainly subtract $\Bbb P\big((A_1\cap A_2')\cap(A_1'\cap A_2)\big)$, but that’s $\Bbb P(\varnothing)=0$, so there’s no need to do so.

Since the events are independent, you then have $$\Bbb P(A_1\cap A_2')=p_1(1-p_2)$$ and $$\Bbb P(A_1'\cap A_2)=(1-p_1)p_2\;,$$ from which the result is a matter of a little algebra.

Solution 1

Clearly, either both events will occur, neither event will occur, or exactly one of the events will occur. The probability of the event will probability $p_1$ happening is, well, $p_1$. The probability of the other event not happening is $1-p_2$. So, the probability that event $1$ occurs and event $2$ does not is $p_1 \cdot (1-p_2)$. Similarly, the probability of event $2$ happening and event $1$ not happening is $p_2 \cdot (1 - p_1)$. Thus, the sum of the two probabilities, which is the probability of exactly one of the events happening, is $$p_1 \cdot (1-p_2) + p_2 (1-p_1) = p_1 + p_2 - 2p_1p_2.$$

Solution 2

We use complementary counting. The probability that neither event occurs is $(1-p_1) \cdot (1-p_2)$ and the probability that both events occur is $p_1 \cdot p_2$. The probability that either both events occur or both events don't occur is $$(1-p_1) \cdot (1-p_2) + p_1 \cdot p_2 = 1 - p_1 - p_2 + 2p_1p_2.$$The probability of the complement is $1 - \text {this}$, so we have: $$1 - \left( 1 - p_1 - p_2 + 2p_1 p_2 \right) = p_1 + p_2 - 2p_1p_2,$$which is the same thing we got with our other solution.

$\blacksquare$

• This is how I'd solve it. I personally don't like fancy unions. This just makes sense. It's the probably of A and not B + the probability of B and not A. – Cruncher Oct 31 '13 at 16:33
• Yeah, the other notation just formally puts it in terms of sets, which is nice, but not required to solve the problem here. Thanks! – Ahaan S. Rungta Oct 31 '13 at 16:47

HINT: Calculate $1-\mathbb{P}(A_1\cap A_2)-\mathbb{P}(A_1^{'} \cap A_2^{'}).$

What do you know about independent events and the independence of their complements?

You need to solve it this way:

$P = p_1 * (1 - p_2) + p_2 * (1 - p_1) = p_1 - p_1p_2 + p_2 - p_1p_2 = p_1 + p_2 - 2p_1p_2$

Explanation: at first we get the probability that $p_1$ will occur and $p_2$ wont. we get: $p_1 * (1 - p_2)$, then we get the probability that $p_2$ will occur and $p_1$ wont: $p_2 * (1 - p_1)$ and add to our first probability.