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
Probability Help!!!
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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
4!/2!*2! * 1/5^2*4/5*2 = 6*1/25*16/25=96/625
nice problem,
take care
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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
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
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
Your solution wud worked if the problem stated "ATLEAST 2"..
hope this clears your point..
Anuu
- force5
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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...
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...