The incident of occupational disease in a industry is such that the workers have a 20% chance of suffering from it what is the probability that out of six workers ?(1)four or more will contact disease ?(2)three will contain disease?
Answers
Answer:
Explanation:
Here 20% of the workers have a chance of suffering from the disease.
Therefore
p= 20/100
= 1/5
( 1- p) = 4/5
. ...(probability of not suffering from the disease).
Now the required probability, where out of 6, none suffers from the disease is
6^Co( 1- p)^6 (p)^0
=(4 /5)^6
The likelihood that three out of every six employees will develop the illness is 0.278.
We can use the binomial distribution to compute the probabilities of various outcomes under the assumption that each worker's likelihood of catching the sickness is independent of the other workers and is 0.2.
(1) The formula for the likelihood that four or more employees would get the disease out of a group of six is P(X 4) = P(X = 4) + P(X = 5) + P(X = 6) where X is the total number of infected employees.
These probability can be determined using the binomial distribution formula as follows:
P(X = 5) = (choose 6) *
P(X = 6) equals (6 choose 6) *
P(X 4) = 0.245 + 0.078 + 0.012 = 0.335 as a result.
Hence, the likelihood that four or more employees out of six will develop the sickness.
(2) The odds that three out of every six employees will develop the illness are as follows: P(X = 3) = (6 pick 3) *
Hence, the likelihood that three out of every six employees will develop the illness is 0.278.
For such more questions on probability,
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