Math question can anyone help? please give a proper answer?

Question

An experimental test to detect a particular type of cancer indicates the presence of cancer in 90% of the induviduals known to have this type of cancer and in 15% of the induviduals known to be cancer free. One hundred induviduals volunteer to take the test. Of the 100, 60 are known to have cancr and 40 are known to be cancer free. If the test indicates that one of the induviduals,randomly chosen has the cancer, what is the probability that he or she is cancer free? thankyou very muc for helping.

Answer 1

Ok, so according to the question, 90% of the people who have the cancer will test positive (and 10% of the people who have the cancer will test negative). That means that 54 of the 60 who actually have the cancer will test positive.

At the same time, 15% of the people who do not have cancer will test positive. That means of the 40 who we know don't have it, 15% will test postive falsely. That means that 6 of the 40 will test positive.

That means out of the entire group of 100, 60 total people will test positive (54 of which will really have it and 6 of which do not).

So if we randomly pick from the 60 who tested positive, there is a 6/60 or 10% chance that the person selected will in actuality NOT have cancer in the first place.

Answer 2

So we want the probability that someone doesn't have it given that they have been shown to have it falsely by an error in the test. Well there is 40*(15/100)=6 people who probabilistically are likely of having that happen to them. And 60*(90/100)=54 people that could have shown to be positive and actually were. So that makes the probability (40*15/100)/(40*15/100+60*90/100)=6/(6+54)=1/10=10%

Answer 3

the answer should be 0% because itthe randomly chosen person would be in the 60 range fo people who had the cancer
it says there are 100 volunteers and 60 opf them have cancer and 40 of them dont so u figure it out there are 60 that have cancer and 40 that dont which means that there is a 0% chance that the 1 person chosen in in the 40 that are cancer free cuz it says that he is known to have the cancer. so do u understand it or did i confuse u even more?

Answer 4

66%
you said cancer was found in 15% of those who were known to be cancer free. Therefore 15% of 40 individuals that are known to be cancer free, have cancer. 15%of 40 is 6 this means 66 individuals out of 100 have cancer which leaves 34% cancer free and 66% with cancer.

Answer 5

First shift all words that have not got "x" to RHS 3x - 2x = 5/2 +a million/4 - 5/3 Now take LCM of two,4 and 3. its 12.Now upload the words on RHS x = (5*6 + a million*3 - 5*4)/12 x = (30 + 3 - 20)/12 x= 13/12 changing into mixed fraction x = a million a million/12

Answer 6

Consider + the event where the test is positive, C where the person has cancer.

From Bayes Theorem: P[C|+] = P[+|C] * P[C] / P[+].

Also,
P[+] = P[+|C] * P[C] + P[+|~C] * P[~C]

P[+|C] is 0.9 (Positive test given cancer is 90%)
P[C] = 0.6 (Probability of cancer is 60%)
P[~C] = 0.4 (Probability of not cancer is 40%)
P[+|~C] = 0.15 (Positive test given not cancer is 15%)

P[+] = 0.9 * 0.6 + 0.15 * 0.4 = 0.6

P[C|+] = 0.9 * 0.6 / P[+] = 0.54 / 0.6 = 0.9

Therefore the probability that a person has cancer given that the test is positive is 90%. And the probability he does not have cancer is 10%

Answer 7

From the experimental tests we know the following.

P(test positive| is really positive) = 0.90
P(test negative| is really positive) = 0.10
P(test positive| is really negative) = 0.15
P(test negative| is really negative) = 0.85

We are told the 0.40 of the individuals is cancer free.

We want to know what the probability that a randomally chosen individual who tests positive is really negative?

0.40*0.15 = 0.06