1
$\begingroup$

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:

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

?

$\endgroup$
  • $\begingroup$ Your first step is not wrong. Just take a closer look at the last term (cf. Brian's answer). $\endgroup$ – caligula Oct 31 '13 at 15:31
3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 $

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Cruncher Oct 31 '13 at 16:33
  • $\begingroup$ Yeah, the other notation just formally puts it in terms of sets, which is nice, but not required to solve the problem here. Thanks! $\endgroup$ – Ahaan S. Rungta Oct 31 '13 at 16:47
2
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.