# By means of an example, show that $P(A) + P(B) = 1$ does not mean that $B$ is the complement of $A$

I'm in grade 10, and I've just started to learn about complementary events. I am rather perplexed with this question. Isn't this question kinda contradictory, since $$P(A) + P(A') = 1$$?

This is what I got to:

$$P(A) + P(B) = 1$$

$$P(A) + P(A') = 1$$

How could it be proven that $$B$$ isn't the complement of $$A$$?

Help would be greatly appreciated.

• I believe a better statement would be "show that $B$ is not necessarily the complement of $A$". – Paras Khosla Mar 24 at 5:49
• A common way to show that something doesn’t hold is to come up with a specific counterexample. – amd Mar 24 at 5:51
• If you flip a coin and get tails, does it mean that the next person you'll meet cannot be a girl? – Eric Duminil Mar 24 at 10:01
• Welcome to MSE; this is a good first question! Here's a brief tutorial on how we format math on this site, like I did in my edit to your question: math.meta.stackexchange.com/questions/5020/… – Robert Howard Mar 24 at 18:23

Take any event of probability $$\frac 1 2$$ and take $$B=A$$.

• And if it helps to give a more concrete example, we can say: I'm going to flip a coin just once. Let $A$ be the event that the flip will be heads, and let $B$ also be the event that the flip will be heads. – aschepler Mar 24 at 18:29

A counterexample

Take a normal 6-sided die.

Let event A be "roll any of the numbers 1, 2, 3 or 4". P(A) = 4/6

Let event B be "roll any of the numbers 1 or 2". P(B) = 2/6

P(A) + P(B) = 4/6 + 2/6 = 6/6 = 1

But B is not the complement of A.

The complement of A is the event "roll any of the numbers 5 or 6".

By this example, we've shown that P(A) + P(B) = 1 does not imply that A and B are complements.

Further

It also seems you have misunderstood the question.

You wrote How could it be proven that B isn't the complement of A?

This is not what you need to prove, and you cannot prove it just by knowing the probabilities. What you need to show is that it isn't always the case. You can show this by giving 1 counterexample, as above.

As commented by @amd you can have many counter examples such as in throwing of dice. You can define any two events A and B. Such that n(A) + n(B)=6. So P(A)+P(B)=1. But A and B need not be necessarily disjoint sets.

From the two statements you obtained (correctly), you can further obtain $$P(A') = P(B)$$. But that does not imply $$A'=B$$. Just as $$x^2 = y^2$$ for reals $$x,y$$ does not imply $$x = y$$.

Another example: $$A = \text{Getting a head on coin }A$$ $$B = \text{Getting a head on coin }B$$

Both have probability $$0.5$$, so $$P(A)+P(B)=1$$, but you could get a head on either, both or neither.

Let $$S$$ be any infinite set and let $$e \in S$$. Now let $$A$$ and $$B$$ be a partition of $$S \setminus \{e\}$$ with both $$A$$ and $$B$$ infinite. Let $$P(C)$$ be the probability of drawing randomly an element of $$S$$ that is in the set $$C$$.

Since $$S = A \cup B \cup \{e\}$$ we have:

$$P(A) + P(B) + P(\{e\}) = 1$$

but also

$$P(\{e\}) = 0$$

since it is finite, thus:

$$P(A) + P(B) = 1$$

For example $$S = \mathbb{N}$$, $$e=\{0\}$$, $$A=\{x \mid x \in \mathbb{N}_0 \wedge x \text{ is even}\}$$ and $$B=\{x \mid x \in \mathbb{N}_0 \wedge x \text{ is odd}\}$$.