Page 1 of 1: Q: Biostatistics: probability of being born

dr_saurabh · Credits: 7235 My Scrapbook

The probability of being born with condition A is 0.1 & the probability of being born with the condition B is 0.5 .
If A & B are Independent , then what is the probability of being born with A OR B OR BOTH ?

The answer given is 0.55 .

How do we tackle such a problem ?

bas · Credits: 444 My Scrapbook

what a Q

drnikhil1 · Credits: 3863 My Scrapbook

when or is given in question, the probability rises..hence it is added

so .5+.1=.6 is the probability of having a or b

now when and is givne in q that means both of these condition should be present so probability decreses... then it is multiplied

nowthe probability of having both a and b is

.5*.1=.05

this is the probability of having a and b both

now we require a or b or both

hence .60+.05=.65

this is the prob of having a or b or both

drnikhil1 · Credits: 3863 My Scrapbook

well nobody on this one?

this above clarification holds true if both conditions r mutually exclusive...now if both conditions r not mutually exclusive, the probability is given by
a+b-ab

so.5+.1-.5*.1=.60-.05=.55 is the probabilty of either a or b if these condition r not mutually exclusive

now the probabilty of a and b both is
a*b=.5*.1=.05

now the probability of having either a or b or a and b both is

.55+.05-.55*.05
.60-.0275=.5725

so the probability of either a or b or a and b both is .5725 if these conditions r not mutually exclusive

and the probabilty of a or b or a and b both is .65 if the conditions r mutually exclusive

any more comments plz

willpower · Credits: 760 My Scrapbook

gr8 explanation nikhil!

dr_saurabh · Credits: 7235 My Scrapbook

thanks

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