HELP!!! please this is my last question for my paper
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The probability of an event is the measurement of the chance of that event's occurrence. The probabilities of considered events are:
- P(At least 8 have the disease) ≈ 0.4378
- P(At most 4 have the disease) ≈ 0.0342
Binomial distributions consist of n independent Bernoulli trials. Bernoulli trials are those trials that end up randomly either on success (with probability p) or on failures( with probability 1- p = q (say))
Suppose we have random variable X pertaining to a binomial distribution with parameters n and p, then it is written as
X = B(n,p)
The probability that out of n trials, there'd be x successes is given by
P(X=x) =
Since 10 people can be either diseased or not and they be so independent of each other (assuming them to be selected randomly) , thus, we can take them being diseased or not as outputs of 10 independent Bernoulli trials.
Let we say
Success= Probability of a diseased person tagged as diseased by the clinic
Failure = Probability of a diseased person tagged as not diseased by the clinic.
Then,
P(Success) = p = 72% = 0.72 (of a single person)
P(Failure) = q = 1-p = 0.28
Let X be the number of people diagnosed diseased by the clinic out of 10 diseased people. Then we have: X ≈ B(n+10,P=0.73)
Calculating the needed probabilities, we get:
a) P(At leased 8 have disease) = P(X≥8) =P(X=8) + P(X=9) + P(X=10)
P(X≥8) =
P(X≥8) ≈ 0.2548 + 0.1456 + 0.0374 ≈ 0.4378
b) P(At most 4 have the disease) = P(X≤4) = P(X=0) + P(X=1)+P(X=2)+P(X=3)+P(X=4)
P (X ≤ 4) =
P (X ≤ 4) = 0.000003 + 0.000076+0.00088+0.00604+0.02719
P (X ≤ 4) = 0.0342
Thus,
The probabilities of considered events are:
- P(At leased 8 have disease) = 0.4378 approx
- P(At most 4 have the disease) = 0.0342 approx
Learn more about binomial distribution here:
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