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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.

Robert Howard
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Gordon Ng
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    I believe a better statement would be "show that $B$ is not necessarily the complement of $A$". – Paras Khosla Mar 24 '19 at 05:49
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    A common way to show that something doesn’t hold is to come up with a specific counterexample. – amd Mar 24 '19 at 05:51
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    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 '19 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: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Robert Howard Mar 24 '19 at 18:23

7 Answers7

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

Kavi Rama Murthy
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  • 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 '19 at 18:29
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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.

user985366
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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.

Tojra
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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.

Robert Howard
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Miriam
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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 $x = y$.

user21820
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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 $B=\{x \mid x \in \mathbb{N}_0 \wedge x \text{ is odd}\}$.

Bakuriu
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0

Suppose that in a finite sample space $\mathcal X$, each elementary event is assigned the same probability. Then every two events $\mathcal U, \mathcal V$ containing the same number of elementary events will be assigned equal probabilities.

More concretely, take $\mathcal X$ comprised of $x_1,x_2,...,x_n$ for some integer $n$, with $\mathrm P(x_1)$ $=\mathrm P(x_2)$$=...= \mathrm P(x_n)=$ $ 1 \over n$. Then any set containing $k$ singletons will have probability $k \over n$.

For this reason, it is not necessary that $\mathrm P(\mathcal U) = \mathrm P(\mathcal V)$ implies that $\mathcal U = \mathcal V$ .

Now if $\mathrm P(\mathcal A) + \mathrm P(\mathcal B) = 1$ then $\mathrm P(\mathcal A') = \mathrm P(\mathcal B)$. But it is not convinced that $\mathcal A' = \mathcal B$.

Vu Nguyen
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