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07 Apr 2011, 10:12
Kudos
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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 only 2 will suffer from disesse x implies only (eactly) 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 = 0.153
Is my approach correct?
I came across this problem in website 'urch'. there was no solution posted, so i wanted to check if my approach is rite. Pls correct me if i'm wrong.
Regards,
Anu
Not sure whats the answer for this problem. But here's my approach:
Out of 4 only 2 will suffer from disesse x implies only (eactly) 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 = 0.153
Is my approach correct?
I came across this problem in website 'urch'. there was no solution posted, so i wanted to check if my approach is rite. Pls correct me if i'm wrong.
Regards,
Anu
Archived Topic
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The one time probability of two infected employees is (0.2)^2 * (0.8)^2 This has to be multiplied with 4C2.
Hence P(two infected) = 4C2 * (0.2)^2 * 0.8^2 = 6 * 0.04 * 0.64= 0.1536 or 15.36%
Hence P(two infected) = 4C2 * (0.2)^2 * 0.8^2 = 6 * 0.04 * 0.64= 0.1536 or 15.36%
07 Apr 2011, 11:41
Kudos
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anuu
Your answer matches mine:
Probability of event occurring k times in n-time sequence could be expressed as:
\(P = C^n_k*p^k*(1-p)^{n-k}\)
\(p=\frac{1}{5}\)
\(1-p=1-\frac{1}{5}=\frac{4}{5}\)
\(P = C^4_2*(\frac{1}{5})^2*(\frac{4}{5})^2= \frac{96}{625}\)
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
