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Probability Help!!!

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Probability Help!!!

by anuu » Thu Apr 07, 2011 9:04 am
Probability of an employee to get infected by a disease X is 20%. What is the probability that out of randomly selected four employees, 2 employees will suffer from disease X?


Not sure whats the answer for this problem. But here's my approach:

Out of 4 ,2 will suffer from disesse x implies only (exactly) 2:

so lets say X & y are the probabilities an employee 'suffering' and 'not suffering' from disease resp:

therefore x = 20/100 = 1/5 and y = 80/100 = 4/5
so we can have XXYY (the employees can be arranged in 4!/2/!2! ways = 6ways)

(1/5*1/5 *4/5*4/5) = 16/625
Since there r 6 ways total prob = 16/625 + 16/625...6times

i.e 96/625 = 0.1536
Is my approach correct?

I came across this problem in website "urch". there was no solotion posted, so i wanted to check if my approach is rite. Pls correct me if i'm wrong.

Regards,
Anu
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by gmat-challenger » Thu Apr 07, 2011 10:18 am
I think you are perfectly right. that's what I got as an answer.

4!/2!*2! * 1/5^2*4/5*2 = 6*1/25*16/25=96/625

nice problem,
take care
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by rohu27 » Thu Apr 07, 2011 10:40 am
why are we calculating the probability for ONLY 2 employees to suffer and assuming the rest do not suffer?
cant it be that 2 WILL suffer implies, 3 may also suffer or all the four?
2 WILL suffer =2 WILL NOT suffer? i dnt get this :(
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by anuu » Thu Apr 07, 2011 11:05 am
The problem states what is the probability that out of 4 employees ,2 will suffer from infection..it coudl be any 2 out of the four.

Your solution wud worked if the problem stated "ATLEAST 2"..

hope this clears your point..

Anuu
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by force5 » Thu Apr 07, 2011 11:40 am
Yes you almost got it correct. Here is my approach..
this concept is called conditional probability.

Here we are selecting 2 employees and the other two will automatically get selected. HOW?

lets start..

there are 4 employees. now we need to select 2 out of 4.
part-1..... how to select 2 out of 4... simple 4C2 = 6

part 2..... how will 2 employees be selected out of 4.

SS
SS (reverse)
SN
NS
NN
NN (reverse)
(S= suffer , N=Not suffer)

Can there be any other combination..... ??.........NO

Now when you pick SS then NN automatically gets picked. (WHY?? because you can only select 2 employees who will suffer from disease as per question)
similarly when you pick SN then ..... NS /SN
when..........SS then .....NN

now probability of Suffer is 1/5 and probability of not suffer is 4/5
hence in each case irrespective of the position of N or S the total probability will be
1/5 * 1/5 * 4/5 * 4/5= 16/625
hence total probability will be 6* 16/625 = 96/625

some books also call it conditional probability of events occurring p times in t sequence..

tCp * (PROB OF EVENT A)^P * (PROB OF EVENT B)^t-p

AGAIN I WOULD RECOMMEND THE LONGER WAY FOR THE STARTERS. you can derive this result from the above solution too.

Hope it helps...
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