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Probability · Venn diagrams & set notation

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Probability from Venn diagrams

Why the denominator is always the total in the universal set — outsiders included — where P(A ∪ B) = P(A) + P(B) − P(A ∩ B) comes from in the picture, and how a "given that" condition shrinks the denominator to one circle.

Probability · Venn diagrams & set notation

Probability from Venn diagrams

Why the denominator is always the total in the universal set — outsiders included — where P(A ∪ B) = P(A) + P(B) − P(A ∩ B) comes from in the picture, and how a "given that" condition shrinks the denominator to one circle.

Why it works

A completed Venn diagram is a picture of the sample space. Every member of the group sits in exactly one region of it — inside both circles, inside just one, or outside both — and nobody is counted twice or left out. That is the whole reason probabilities can be read straight off it: the regions are outcomes that cannot overlap and that between them cover everything.

Here are 3030 students, asked whether they play football (FF) or tennis (TT).ξ3FT1467Because each student is in exactly one region, the four numbers must add to the size of the whole group:

14+6+7+3=30.14 + 6 + 7 + 3 = 30.

It works ✓. Do that check first, every time: if the regions don't add to the stated total, something is wrong before a single probability has been worked out.

The denominator is the universal set — outsiders included. The rectangle around the circles is ξ\xi, the universal set: everybody who could be chosen. A student is picked at random from the whole group, so all 3030 are equally likely and the total is 3030. The FF circle holds 14+6=2014 + 6 = 20 students, so

P(F)=2030=23.P(F) = \frac{20}{30} = \frac{2}{3}.

The classic error is to use 2727 — the number inside the circles — and write 2027\frac{20}{27}. Those 33 students who play neither sport are still in the room; they can still be the one picked. Dropping them from the denominator makes every probability on the page too big, and it breaks the one rule probability cannot break. The four regions are the four possible things that can happen to the chosen student, so their probabilities must add to exactly 11. Over 2727 they give

1427+627+727+327=3027>1,\frac{14}{27} + \frac{6}{27} + \frac{7}{27} + \frac{3}{27} = \frac{30}{27} > 1,

which is impossible; over 3030 they give 14+6+7+330=3030=1\frac{14 + 6 + 7 + 3}{30} = \frac{30}{30} = 1 ✓. The outside region is also the only place P(neither)=330=110P(\text{neither}) = \frac{3}{30} = \frac{1}{10} can come from — ignore it and that question becomes unanswerable.

P(A)P(A) is the whole circle, not the crescent. Same picture, second trap: P(F)P(F) is 2030\frac{20}{30}, not 1430\frac{14}{30}. The 1414 is football only. The 66 in the overlap play football as well — they just play tennis too — and they are inside the FF boundary, so they count. Sweep up every number inside the circle you want.

Why you subtract the overlap. For P(F∪T)P(F \cup T) — football or tennis or both — count everyone in at least one circle: 14+6+7=2714 + 6 + 7 = 27, so P(F∪T)=2730=910P(F \cup T) = \frac{27}{30} = \frac{9}{10}.

Now try building the same thing out of P(F)P(F) and P(T)P(T). The TT circle holds 6+7=136 + 7 = 13, so P(T)=1330P(T) = \frac{13}{30}, and

2030+1330=3330,\frac{20}{30} + \frac{13}{30} = \frac{33}{30},

which is bigger than 11 and so cannot be a probability at all. The excess is 33−27=633 - 27 = 6 students — exactly the overlap. Of course it is: the 66 are inside FF and inside TT, so adding the two circles counts them twice, when they should be counted once. Take one copy back off:

P(A∪B)=P(A)+P(B)−P(A∩B).P(A \cup B) = P(A) + P(B) - P(A \cap B).

That is not a formula to memorise, it is a double count being repaired. Here 3330−630=2730\frac{33}{30} - \frac{6}{30} = \frac{27}{30}. ✓ And if the circles don't overlap, P(A∩B)=0P(A \cap B) = 0 and the rule collapses to P(A)+P(B)P(A) + P(B) — which is why the "just add them" method works sometimes and quietly fails whenever there is an overlap.

Conditional probability: the denominator shrinks. "Given that the student plays tennis, find the probability they also play football." That word given changes who could have been picked. Not all 3030 — only the 1313 inside the TT circle. Rebuild the fraction inside that circle: of those 1313, the ones who also play football are the 66 in the overlap. So

P(F∣T)=613.P(F \mid T) = \frac{6}{13}.

The TT circle has become the new universal set. Notice 613\frac{6}{13} is not 630\frac{6}{30} and not 2030\frac{20}{30} — being told something genuinely changes the number, because it rules people out.

Mutually exclusive and independent are two different tests.
  • Mutually exclusive — they cannot both happen — means the overlap is empty,
P(A∩B)=0P(A \cap B) = 0. You can see this: look at the intersection region.
  • Independent — knowing one happened doesn't change the chance of the other
— means P(A∩B)=P(A)×P(B)P(A \cap B) = P(A) \times P(B) (equivalently P(A∣B)=P(A)P(A \mid B) = P(A)). You cannot see this; you must calculate it.

For the students: the overlap holds 66, so FF and TT are not mutually exclusive. And P(F)×P(T)=2030×1330=260900≈0.289P(F) \times P(T) = \frac{20}{30} \times \frac{13}{30} = \frac{260}{900} \approx 0.289, while P(F∩T)=630=0.2P(F \cap T) = \frac{6}{30} = 0.2. These are not equal, so FF and TT are not independent either. Both tests can fail at once — the two words are not opposites, and answering one of them never answers the other.