H 333 (A): Probability & Statistics. Examination #1 (Fall 2008) 3. Of patients g
ID: 3329368 • Letter: H
Question
Explanation / Answer
Here we are given that P( specialist )= 0.3 , P( lab work ) = 0.4 and P( neither of the two ) = 0.35
a) Now as we are given that: P( neither of the two ) = 0.35, therefore
P( specialist or lab work ) = 1 - P( neither of the two ) = 1 - 0.35 = 0.65
Using addition law of probability, we get:
P( specialist and lab work ) = P( specialist ) + P( lab work ) - P( specialist or lab work )
P( specialist and lab work ) = 0.3 + 0.4 - 0.65 = 0.05
Therefore 0.05 is the required probability here.
b) Probability that all 5 of them are referred to a specialist is computed as:
= 0.3*0.3*0.3*0.3*0.3
= 0.35
= 0.00243
Therefore 0.00243 is the required probability here.
c) Probability that at least one of the next 5 is referred to the specialist
= 1 - Probability that no one is referred to the specialist
= 1 - (1 - 0.3)5
= 0.83193
Therefore 0.83193 is the required probability here.
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