Thread: binomial random variable Expectation

Each of 1000 people in a place independently has a certain
disease with probability 0.005.

The disease can be show up in the blood test by two methods
Method 1- Each person has separate test
Method 2- Ten groups of 100 people are formed. The blood samples of 100 people in a group are mixed and later tested. Given that test is positive, implying that at least one person in the group has the disease, all 100 people must be tested separately. We do this for all the ten groups.
(i). What is the expected number of blood tests performed for the 1000 people in method 1? Is it 0.005 x 1000=5
(ii)In method 2, if y is the number of blood tests performed in a single group of 100 people, what is the expected value of y?

Originally Posted by someone21

Each of 1000 people in a place independently has a certain
disease with probability 0.005.

The disease can be show up in the blood test by two methods
Method 1- Each person has separate test
Method 2- Ten groups of 100 people are formed. The blood samples of 100 people in a group are mixed and later tested. Given that test is positive, implying that at least one person in the group has the disease, all 100 people must be tested separately. We do this for all the ten groups.
(i). What is the expected number of blood tests performed for the 1000 people in method 1? Is it 0.005 x 1000=5
(ii)In method 2, if y is the number of blood tests performed in a single group of 100 people, what is the expected value of y?

For (i) the expected number of tests is 1000 because every person is tested. Surely you mean the expected number of blood tests that show positive, in which case5 is the correct answer.

What have you tried for (ii)?

Originally Posted by mr fantastic

For (i) the expected number of tests is 1000 because every person is tested. Surely you mean the expected number of blood tests that show positive, in which case5 is the correct answer.

What have you tried for (ii)?

for i, the question just asks that so I believe that you are correct althought then i cannot believe it is so simple

for ii, I am just unable to find the way to do

Lastly why cannot be use the binomial expectation formula

Originally Posted by someone21

for i, the question just asks that so I believe that you are correct althought then i cannot believe it is so simple

for ii, I am just unable to find the way to do

Lastly why cannot be use the binomial expectation formula Mr F says: You can. What makes you think you can't?

i. np = (1000)(0.005) = 5.

ii. np = (100)(0.005) = 0.5.