Part II ProbabilityProbability
3.4 Conditional Probability
3.4.1 Evaluating Conditional Probabilities
To get an expression forP.BjA/, notice that, becauseAis known to have occurred, our space of outcomes or sample space is now reduced toA. We know that our outcome lies inA;P.BjA/is the probability that it also lies inB\A.
The outcome lies inA, and so it must lie in eitherB\Aor inBc\A, and it cannot lie in both. This means that P.BjA/CP.BcjA/D1:
Now recall the idea of probabilities as relative frequencies. IfP.C\A/D kP.B\A/, this means that outcomes inC\A will appearktimes as often as outcomes inB\A. But this must apply even if we know in advance that the outcome is inA. This means that, ifP.C\A/DkP.B\A/, thenP.CjA/DkP.BjA/. In turn, we must have
P.BjA//P.B\A/:
Now we need to determine the constant of proportionality; writecfor this constant, meaning P.BjA/DcP.B\A/:
We have that
P.BjA/CP.BcjA/DcP.B\A/CcP.Bc\A/DcP.A/D1;
so that
P.BjA/D P.B\A/ P.A/ :
I find the “size” metaphor helpful here. We have thatP.BjA/measures the probability that an outcome is inB, given we knowit is inA. From the “size” perspective,P.BjA/measures the “size” of.A\B/relative toA. So our expression makes sense, because the fraction of the eventAthat is also part of the eventBis given by the “size” of the intersection divided by the “size” ofA.
Another, very useful, way to write the expressionP.BjA/DP.B\A/=P.A/is:
P.BjA/P.A/DP.B\A/:
Now, sinceB\ADA\B, we must have that
P.BjA/D P.AjB/P.B/ P.A/
Worked example 3.26 (Car Factories) There are two car factories,A andB. Each year, factory Aproduces 1000 cars, of which 10 are lemons. FactoryBproduces 2 cars, each of which is a lemon. All cars go to a single lot, where they are thoroughly mixed up. I buy a car.
• What is the probability it is a lemon?
• What is the probability it came from factoryB?
• The car is now revealed to be a lemon. What is the probability it came from factoryB, conditioned on the fact it is a lemon?
(continued)
68 3 Basic Ideas in Probability
Solution
• Write the event the car is a lemon asL. There are 1002 cars, of which 12 are lemons. The probability that I select any given car is the same, so we haveP.L/D12=1002.
• Same argument yieldsP.B/D2=1002.
• WriteBfor the event the car comes from factoryB. I needP.BjL/DP.L\B/=P.L/DP.LjB/P.B/=P.L/. I have P.LjB/P.B/=P.L/D.12=1002/=.12=1002/D1=6.
Worked example 3.27 (Royal Flushes in Poker—1) You are playing a straightforward version of poker, where you are dealt five cards face down. A royal flush is a hand of AKQJ10 all in one suit. What is the probability that you are dealt a royal flush?
Solution This is
number of hands that are royal flushes, ignoring card order total number of different five card hands, ignoring card order:
There are four hands that are royal flushes (one for each suit). Now the total number of five card hands is 52
5
D2;598;960
so we have
4
2;598;960 D 1 649;740:
Worked example 3.28 (Royal Flushes in Poker—2) You are playing a straightforward version of poker, where you are dealt five cards face down. A royal flush is a hand of AKQJ10 all in one suit. The fifth card that you are dealt lands face up. What is the conditional probability of getting a royal flush, conditioned on the event that this card is the nine of spades?
Solution No hand containing a nine of spades is a royal flush, so this is easily zero.
Worked example 3.29 (Royal Flushes in Poker—3) You are playing a straightforward version of poker, where you are dealt five cards face down. A royal flush is a hand of AKQJ10 all in one suit. The fifth card that you are dealt lands face up. It is the Ace of spades. What now is the probability that your have been dealt a royal flush? (i.e. what is the conditional probability of getting a royal flush, conditioned on the event that one card is the Ace of spades)
Solution Now consider the events
ADyou get a royal flushandthe last card is the ace of spades
and
BDthe last card you get is the ace of spades;
(continued)
and the expression
P.AjB/D P.A\B/
P.B/ : NowP.B/D 521.P.A\B/is given by
number of five card royal flushes where card five is Ace of spades total number of different five card hands : This is
4321 5251504948 yielding
P.AjB/D 1 249;900: Notice the interesting part: seeing this card has really made a difference.
Worked example 3.30 (Two Dice) We throw two fair six-sided dice. What is the conditional probability that the sum of spots on both dice is greater than six, conditioned on the event that the first die comes up five?
Solution Write the event that the first die comes up 5 asF, and the event the sum is greater than six asS. There are five outcomes where the first die comes up 5 and the number is greater than 6, soP.F\S/D5=36. Now
P.SjF/DP.F\S/=P.F/D.5=36/=.1=6/
D5=6:
Notice thatA\BandA\Bcare disjoint sets, and thatAD.A\B/[.A\Bc/. So, becauseP.A/DP.A\B/CP.A\Bc/, we have
P.A/DP.AjB/P.B/CP.AjBc/P.Bc/
a tremendously important and useful fact. Another version of this fact is also very useful. Assume we have a collection of disjoint setsBi. These sets must have the property that (a)Bi\Bj D ¿fori ¤ jand (b) they coverA, meaning that A\.[iBi/DA. Then, becauseP.A/DP
iP.A\Bi/, so we have P.A/DX
i
P.AjBi/P.Bi/
It is wise to be suspicious of your intuitions when thinking about problems in conditional probability. There is a really big difference betweenP.AjB/andP.BjA/P.A/. Not respecting this difference can lead to serious problems (Sect.3.4.4), and seems to be easy to do. The division sign in the expression
P.AjB/DP.BjA/P.A/=P.B/
can have alarming effects; as a result, most people have quite poor intuitions about conditional probability.
Remember this:Here is one helpful example. If you buy a lottery ticket (L), the probability of winning (W) is small.
So P.WjL/may be very small. But P.LjW/is 1—the winner is always someone who bought a ticket.
70 3 Basic Ideas in Probability
Useful Facts 3.3 (Conditional Probability Formulas) You should remember the following formulas:
• P.BjA/D P.AjB/P.B/ P.A/
• P.A/DP.AjB/P.B/CP.AjBc/P.Bc/
• Assume (a)Bi\BjD¿fori¤jand (b)A\.[iBi/DA; thenP.A/DP
iP.AjBi/P.Bi/