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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
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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.
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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= = =
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