counting probablity 3 conditional

8/13/2019 Counting Probablity 3 Conditional

1/5

Conditional Probability

Recap: Independent events.

If events A and B are independent, then the probability of B happening does not depend upon

whether A has happened or not. Therefore

P(B | A) P(B | A )=P(B)=

and

P(A and B) = P(A) P(B) .

A tree diagra !an be drawn"

Dependent events:

If A and B are dependentevents, the probability of B happening will depend upon whether A has

happened or not.

#e therefore have to introdu!e !onditional probabilities in the tree diagra (as shown below)"

Probability of Bgiven that A has

o!!urred


2/5


3/5

otation" T = buy The Ties ' = buy The 'ail / = !oplete !rossword

a) -ro the tree diagra,

P(!oplete !rossword) = P(T and /) 0 P(' and /) =1 2

*3 *3 *+ =

b) The probability that I bought The 'ail given that I !opleted the !rossword is given by

P(' and /) **3P(' | /) =

2P(/) +*

= =

Example 2:

3.24 of the population !arry a parti!ular faulty gene.

A test e5ists for dete!ting whether an individual is a !arrier of the gene.In people who a!tually !arry the gene, the test provides a positive result with probability 3.6.

In people who don7t !arry the gene, the test provides a positive result with probability 3.32.

If soeone gives a positive result when tested, find the probability that they a!tually are a !arrier of

the gene.

8et 9 = person !arries gene P = test is positive for gene = test is negative for gene

The tree diagra then loo:s as follows"

#e want to findP(9 and P)

P(9 | P) =P(P)

;owever, P(P) = P(9 and P) 0 P(9< and P) = 3.3336 0 3.33666 = 3.3236

Therefore,3.3336

P(9 | P) = 3.3*13.3236

= (to > signifi!ant figures)

?o there is a very low !han!e of a!tually having the gene even if the test says that you have it.

ote" This e5aple highlights the diffi!ulty of dete!ting rare !onditions or diseases.


4/5


5/5

Past Examination Question: (OCR

?tudents have to pass a test before they are allowed to wor: in a laboratory. ?tudents do not reta:e

the test on!e they have passed it. -or a randoly !hosen student, the probability of passing the test

at the first attept is 2@>. n any subse%uent attept, the probability of failing is half the

probability of failing on the previous attept. By drawing a tree diagra or otherwise,

a) show that the probability of a student passing the test in > attepts of fewer is *1@*,

b) find the !onditional probability that a student passed at the first attept, given that the studentpassed in > attepts or fewer.

!olution:

8et P = pass test - = fail test

a) ?o P(pass in > attepts or fewer) = 2@> 0 @6 0 +@* = *1@* as re%uired

b) 8et Xstand for when a student passes the test.

#e needP( 2 and >)

P( 2| >)P( >)

X XX X

X

= = =

.

But P(X= 2 and XC >) is the sae as Dust P(X= 2).

Therefore2>

*1*B

6P( 2| >)

*1X X= = =

counting probablity 3 conditional

Documents