Combined Events

Naming the Regions of a Venn Diagram #

The overlap, each circle alone, and the box.

Two circles cut everything into the overlap, each circle alone, and the box outside both

These 18 homes are sorted by whether they keep a dog, a cat, both, or neither.

The overlap is the 3 homes that keep a dog and a cat at once.

The dog circle without the overlap is the 6 homes that keep a dog and no cat.

The box outside both circles is the 4 homes that keep neither pet.

Two symbols name these regions: $A\cap B$A ∩ B is the overlap, $A\cup B$A ∪ B is either circle.

Possibility Diagrams #

Every pair as a dot, so none get missed.

A grid holds every possible pair, so no outcome gets left out

One dice runs along the bottom and the other up the side. Each of the 36 dots is one pair.

Every dot is one outcome. These 5 dots are the pairs that total 6.

Chance from Areas #

Target area over total area.

When a dart is equally likely to land anywhere, the chance is target area over total area

A dart lands somewhere on this board, and every spot is equally likely.

The target is a 2-by-2 patch. It covers 4 of the 20 equal squares.

Chance is target area over total area: 4 squares out of 20 is $\frac{1}{5}$1/5.

Mutually Exclusive Events #

Either one or the other, so add the chances.

Mutually exclusive events cannot both happen, so their chances simply add

One roll cannot show a 1 and a 6 at once, so the overlap holds nothing: 0.

Of the 6 faces, 1 shows a 1 and 1 shows a 6. The other 4 sit outside both circles.

The two events share no face, so nothing is subtracted: $\frac{1}{6}+\frac{1}{6}=2$1/6 + 1/6 = 2 of the 6 faces.

The Addition Rule for Overlapping Events #

Add the chances, then subtract the shared part.

Adding two overlapping chances counts the shared outcomes twice, so subtract them once

A dice has 6 faces. 3 of them are even, 2 are multiples of 3, and the face 6 is both.

Add $\frac{3}{6}$3/6 and $\frac{2}{6}$2/6 and the face showing 6 has been counted twice.

Subtract that shared face once: 3 + 2 − 1 leaves 4 of the 6 faces.

So add both chances, then subtract the chance of both happening.

Independent Events #

One does not affect the other, so multiply.

When two events are independent, multiply their chances to get the chance of both

A dice and a coin are independent: neither one changes the chances on the other.

Lay out every pair instead: each of the 6 faces with each of the 2 coin sides, all equally likely.

Just one of the 12 pairs is a six with a head, so its chance is 1 in 12.

The 12 is 6 × 2, so $\frac{1}{12}$1/12 is $\frac{1}{6}\times \frac{1}{2}$1/6 × 1/2. That is why the chances multiply.

Tree Diagrams #

Multiply along a path through the tree.

A tree diagram shows every path through two stages at once

Two coins are tossed. Each branch splits again, so the tree has four paths.

Multiply the chances along a path: $\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$1/2 × 1/2 = 1/4 for two heads.

The Chance of At Least One #

One minus the chance of none at all.

The chance of at least one success is one minus the chance of no successes at all

Two rolls of a dice make 36 outcomes. Every square that holds a six is marked.

In only these 25 outcomes does neither roll show a six: $\frac{5}{6}\times \frac{5}{6}=\frac{25}{36}$5/6 × 5/6 = 25/36.

At least one six is everything that is left over: 36 − 25 = 11 outcomes.

The chance of at least one is 1 minus the chance of none, however many tries there are.

Without Replacement #

Taking one out changes the next chance.

Taking a counter out and not replacing it changes the chances on the next draw

The bag holds three red counters and two blue. Take one out and four are left.

The second branch reads $\frac{2}{4}$2/4, not $\frac{3}{5}$3/5, because the bag has changed.

Multiplying Along a Tree Without Replacement #

Multiply along a path; add the paths.

Multiply the chances along a path, then add the paths that cannot both happen

The same bag of three red and two blue. Red then red is one path through the tree.

Multiply along the path: $\frac{3}{5}$3/5 of draws start red, and $\frac{2}{4}$2/4 of those draw red again.

One of each color can happen two ways. Red then blue is $\frac{3}{5}\times \frac{2}{4}=\frac{3}{10}$3/5 × 2/4 = 3/10.

Blue then red is the other way of getting one of each: $\frac{2}{5}\times \frac{3}{4}=\frac{3}{10}$2/5 × 3/4 = 3/10.

The two paths are mutually exclusive, so add their chances: $\frac{3}{10}+\frac{3}{10}=\frac{3}{5}$3/10 + 3/10 = 3/5.

Conditional Probability #

Knowing shrinks what is still in play.

Knowing one event has happened cuts down the outcomes that are still possible

These 24 people are sorted by whether they play music, sport, both, or neither.

Given that a person plays music, only these 12 count, and 5 of them also play sport.

Here is a rare illness and a test that is right most of the time. Every yes is shaded.

Given that the test said yes, only that strip counts, and nearly all of it is well people.

Conditional Probability Notation #

A vertical bar reads as the word given.

A vertical bar reads as given, and the chance is the shared part divided by the condition

P(A | B) is read as the chance of A, given that B has happened.

Given B, only the 13 outcomes inside B count: 5 shared with A and 8 in B alone.

Of those 13, the 5 in the overlap are in A as well, so P(A | B) is $\frac{5}{13}$5/13.

So P(A | B) is the chance of both, divided by the chance of B.

The Monty Hall Problem #

Switching wins two times in three.

Switching doors wins two times in three, because the host never opens the car

There are three doors and one car. Your first pick is right one time in three.

The host knows where the car is and always opens a door with a goat.

If you picked a goat, the host must open the other goat, so the last door hides the car.

So the $\frac{2}{3}$2/3 chance that your first pick was wrong now belongs to the other door.

The Birthday Paradox #

Twenty-three people, and evens on a match.

In a room of twenty-three people, a shared birthday is more likely than not

5 people make 10 pairs, and 23 people make 253. Any one of those pairs could match.

Each of the 10 pairs matches about 1 day in 365, so about 3 rooms in 100 have a match.

The other 97 rooms in 100 are the ones where nobody matches: 100 − 3 = 97.

The curve passes one half at 23 people, and reaches 99 percent by 57.

Why a Positive Test Can Still Mean Healthy #

When it is rare, most positives are false.

When an illness is rare, most positive tests come from the healthy majority

1 person in 1000 has this illness, and the test is right 99 times in 100.

Of a thousand people, one is ill and about ten healthy people also test positive.

Of the 11 people who test positive, only 1 is really ill. The other 10 are false alarms.

Reversing a Tree: the Base-Rate Answer #

Ill-and-positive over every positive.

Reading a tree diagram backwards divides one path by every path that ends the same way

The test says positive. Two paths end there: ill and correctly caught, or well and misread.

Add the two paths that end positive: about 11 people in a thousand test positive.

Given only that the test said yes, the top strip is all that counts, and nearly all of it is well.

Divide the ill path by all the positives: 0.00099 ÷ 0.01098, which is about 9 percent.

So a positive from a test that is right 99 times in 100 means ill about 1 time in 11.