Respuesta :
Answer: 10773
Step-by-step explanation:
.75^10=0.0563
look at 8, 9, 10
8 the probability is 10C8*0.75^8*0.25^2=0.2816
9 the probability is 0.1877 using the same approach
At least 8 is 0.5256
At most 4 is
0 .25^10=essentially 0
1: 10C1*0.75*0.25^9=<0.0001
2: 45*0.75^2*0.25^8=0.0004
3: 120*0.75^3*0.25^7=0.0031
4: 10C4*0.75^4*0.25^6=0.0162
That sum is 0.0197
Probability of an event is the measurement of its chance of that event's occurrence. The probabilities of considered events are:
- P(At leased 8 have disease) ≈ 0.4378
- P(At most 4 have the disease) ≈ 0.0342
How to find that a given condition can be modeled by binomial distribution?
Binomial distributions consists of n independent Bernoulli trials.
Bernoulli trials are those trials which 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 binomial distribution with parameters n and p, then it is written as
[tex]X \sim B(n,p)[/tex]
The probability that out of n trials, there'd be x successes is given by
[tex]P(X =x) = \: ^nC_xp^x(1-p)^{n-x}[/tex]
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 clinic out of 10 diseased people. Then we have: [tex]X \sim B(n=10, p=0.72)[/tex]
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) = [tex]^{10}C_8(0.72)^8(0.28)^2 + \: ^{10}C_9(0.72)^9(0.28)^1 + \: ^{10}C_{10}(0.72)^{10}(0.28)^0[/tex]
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)
or
[tex]P(X \leq 4) = \: ^{10}C_0(0.72)^0(0.28)^{10} + \: ^{10}C_1(0.72)^1(0.28)^9 + \: ^{10}C_2(0.72)^2(0.28)^8 + \: ^{10}C_3(0.72)^3(0.28)^7 + \: ^{10}C_4(0.72)^4(0.28)^6 \\\\P(X \leq 4) \approx 0.000003 + 0.000076 + 0.00088 + 0.00604 + 0.02719\\\\P(X \leq 4) \approx 0.0342[/tex]
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:
https://brainly.com/question/13609688
