# Conditional Probability

The probability of 7 when rolling two die is 1/6 (= 6/36) because the sample space consists of 36 equiprobable elementary outcomes of which 6 are favorable to the event of getting 7 as the sum of two die. Denote this event A:

Consider another event B which is having at least one 2. There are still 36 elementary outcomes of which 11 are favorable to B; therefore,

The assumption that B took place reduces the set of possible outcomes to 11. Of these, only two - 25 and 52 - are favorable to A. Since this is reasonable to assume that the 11 elementary outcomes are equiprobable, the probability of A under the assumption that B took place equals 2/11. This probability is denoted *conditional probability* of A assuming B. Of course, for any event A, *universal event* - the whole of the sample space - for which all available elementary outcomes are favorable.

We see that, in our example,

Retracing the steps in the example,

P(A|B) = P(A∩B) / P(B),

and this is the common definition of the *conditional probability*. The formula confirms to the earlier definitions:

P(A|Ω) = P(A∩Ω) / P(Ω) = P(A)

since P(Ω) = 1 and

For another example, let's look at the events associated with rolling a dice in the form of octahedron, a shape with eight faces. The sample space naturally consists of 8 equiprobable outcomes:

Ω = {1, 2, 3, 4, 5, 6, 7, 8}.

Let A be the event of getting an odd number, B is the event getting at least 7. Then

P(A|B) = P(A∩B)/P(B) = 1/2 = P(A).

Observe also that

P(B|A) = P(A∩B)/P(A) = 1/4 = P(B).

On the other hand, define

A_{+} = {1, 2, 3, 5, 7} and

A_{-} = {3, 5, 7}.

Then

P(B|A_{+}) = P(A_{+}∩B)/P(A_{+}) = 1/5 < 1/4 = P(B),

whilst

P(B|A_{-}) = P(A_{-}∩B)/P(A_{-}) = 1/3 > 1/4 = P(B).

So we see that, in general, there is no definite relationship between the probability P(B) and the conditional probability P(B|A). They may be equal, or one of them may be greater than the other. In the former case the events are said to be *independent*.

Conditional probabilities arise naturally (and prove useful) in the investigation of experiments run repeatedly where an outcome of a trial may affect the outcomes of the subsequent trials. *Drawing without replacement* is one class of such experiments. Making a selection by drawing a straw or a match from a bunch of apparently equal items is a common practice throughout the world [Falk, 2.3.3].

For a group of six kids, six identical-looking matches are used. An uninvolved person secretly breaks one match at its lower end and then holds all six in his/her palm, so that the lower ends are hidden, and only six upper ends are visible arranged evenly.

The first child randomly draws a match. If the match is the broken one, the child is selected, and the drawing procedure stops. If not, the second child draws a match, and so on. The drawing procedure terminates when the broken match is selected.

Let us denoted the six children by their ordinal numbers in the draw: 1, 2, ..., 6. (The randomness, or rather arbitrariness of the order can be assured by an entertaining activity of drawing shuttles.) For each child, compute the probability that he or she will be the selected person.

The probability for the first child is obviously 1/6 because there are 6 matches with equal chances of being drawn. The probability that the first child will not be selected is 5/6. If it's not, the remaining 5 matches have equal chances to be drawn by the second child who, therefore, has the probability of 1/5 of being selected on the second drawing. But this probability is conditional on the failure of the first child to win the selection. In other words, the conditional probability

P(B) | = P(B∩A) + P(B∩A) |

= P(B∩A) | |

= P(B|A)×P(A) | |

= 1/5 × 5/6 | |

= 1/6. |

This is because P(A∩B) = 0, as only one child could be selected. All individual child selections are mutually exclusive events.

The probability that neither of the first two children is selected

P(C) | = P(C∩(A∩B)) |

= P(C|A∩B)×P(A∩B) | |

= 1/4 × 4/6 | |

= 1/6. |

The similar reasoning applies to the remaining children so that all six of them have an equal chance to draw the broken match. The procedure is quite fair, although, at first sight, the result is counterintuitive.

In describing the survival rate and life expectancy in a certain population, let A_{N} denote the event of reaching the age of N years and _{N})

P(50) = .913,

P(55) = .881,

P(65) = .746.

This information suggests several questions. For example, what is the probability of a 50 years old man to reach the age of 55, i.e. what is _{55}|A_{50})?_{55}∩A_{50} = A_{55},

P(55|50) = P(A_{55}∩A_{50})/P(A_{50}) = P(A_{55})/P(50) ≈ .965.

A probability that a 50 years old will die within 5 years is then a rather comforting ^{th} birthday, i.e., what is P(70)?

As before, P(70|65) = P(70)/P(65) so that P(70) = P(65)·P(70|65), but

P(70|65) = 1 - .16 = .84.

Therefore,

P(70) = P(65)·P(70|65) = .746·.84 ≈ .627.

**Note:** An interactive tool to check your understanding of conditional probabilities is available at https://www.cut-the-knot.org/Curriculum/Probability/Conditionalprobability.shtml.

### References

- R. B. Ash,
*Basic Probability Theory*, Dover, 2008 - R. Falk,
*Understanding Probability and Statistics*, A K Peters, 1993

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