Expected values probability help

[puta]

A 20-year-old male purchases a 1-year life insurance policy worth 250,000 dollars. The insurance company determines that he will survive the policy period with probability 0.9986.

(a) If the premium for the policy is 410 dollars, what is the expected profit for the insurance company?

i did $410* 0.9986- *250,000*0.0014
but i don't think it's right?
b) At what value should the company set its premium so its expected profit will be 240 dollars per policy for 20-year-old males?

and i am not really sure how to approach this oneand also
A box contains 3 defective bulbs and 9 good bulbs. If 5 bulbs are drawn from the box without replacement, what is the expected number of defective bulbs?

i did 5 *(3/9) what else do i need to consider?

 

[HallsofIvy]

A 20-year-old male purchases a 1-year life insurance policy worth 250,000 dollars. The insurance company determines that he will survive the policy period with probability 0.9986.

(a) If the premium for the policy is 410 dollars, what is the expected profit for the insurance company?

i did $410* 0.9986- *250,000*0.0014
but i don't think it's right?

It's almost right. If they person survives the company makes $410. If the person does not, the company pays out $250000 but they still have the origina $410. The net loss to the company if the person dies is $250- 410. 410(0.9986)- (250000- 410)(0.0014)= 410(0.9986+ 0.0014)- 250000(0.0014)= 410- 250000(.0014).

b) At what value should the company set its premium so its expected profit will be 240 dollars per policy for 20-year-old males?

and i am not really sure how to approach this one

Replace the premium given above by the variable P in that same formula: P- 250000(0.0014), set it equal to $240, and solve for P.

and also
A box contains 3 defective bulbs and 9 good bulbs. If 5 bulbs are drawn from the box without replacement, what is the expected number of defective bulbs?

i did 5 *(3/9) what else do i need to consider?

n*p is the expected value for "sampling with replacement". Without replacement, to find the probability getting "DGGGG" (where D is a defective bulb and G a good bulb), the probabilty the first bulb draw is defective is 3/12, there are then 11 bulbs left, 9 of then good so the probability the second bulb is good is 9/11, the probability the third bulb is good is 8/10, the probability that the fourth bulb is good is 7/9, and the probability the fifth bulb is good is 6/8. The probability of "DGGGG" (in that order) is (3/12)(9/11)(8/10)(7/9)(6/8). But there are allso 5 different possible orders: "DGGGG", "GDGGG", "GGDGG", "GGGDG", and "GGGGD" so the probability of exactly one defective bulb is 5(3/12)(9/11)(8/10)(7/9)(6/8).

The probability of "DDGGG" is (3/12)(2/11)(9/10)(8/9)(7/8) and there are [itex]_5C_2= 5!/3!2!= 10 such orders: the probability of two defective bulbs is 10(3/12)(2/11)(9/10)(8/9)(7/8).

Similarly, find the probability of 3, 4, and 5 defective bulbs. The "expected value" is the sum of the products of the number of defective bulbs times the probability of that number.

 

[Ray Vickson]

His expected value is correct, but probably not for the reasons he thinks. The number of detectives in his sample will follow the hypergeometric distribution, whose exact mean happens to be given by the formula he used. An alternative derivation is to note that the number of defectives, N, is N = sum{ I_j : j = 1...5}, where I_j is an indicator variable for the jth draw being defective or not. We have EN = sum{EI_j : j = 1...5}, and each I_j has the same expected value.

RGV

 

[puta]

It's almost right. I

n*p is the expected value for "sampling with replacement". Without replacement, to find the probability getting "DGGGG" (where D is a defective bulb and G a good bulb), the probabilty the first bulb draw is defective is 3/12, there are then 11 bulbs left, 9 of then good so the probability the second bulb is good is 9/11, the probability the third bulb is good is 8/10, the probability that the fourth bulb is good is 7/9, and the probability the fifth bulb is good is 6/8. The probability of "DGGGG" (in that order) is (3/12)(9/11)(8/10)(7/9)(6/8). But there are allso 5 different possible orders: "DGGGG", "GDGGG", "GGDGG", "GGGDG", and "GGGGD" so the probability of exactly one defective bulb is 5(3/12)(9/11)(8/10)(7/9)(6/8).

The probability of "DDGGG" is (3/12)(2/11)(9/10)(8/9)(7/8) and there are [itex]_5C_2= 5!/3!2!= 10 such orders: the probability of two defective bulbs is 10(3/12)(2/11)(9/10)(8/9)(7/8).

Similarly, find the probability of 3, 4, and 5 defective bulbs. The "expected value" is the sum of the products of the number of defective bulbs times the probability of that number.

i understood the explanation but like is that the only way to do it? because if it's on a test wouldn't that take a very long time to solve for the answer?

thank you for the help, i really appreciated

 

[puta]

His expected value is correct, but probably not for the reasons he thinks. The number of detectives in his sample will follow the hypergeometric distribution, whose exact mean happens to be given by the formula he used. An alternative derivation is to note that the number of defectives, N, is N = sum{ I_j : j = 1...5}, so EN = sum{EI_j : j = 1...5}, and each I_j has the same expected value.

RGV

N, is N = sum{ I_j : j = 1...5}, so EN = sum{EI_j : j = 1...5}, and each I_j has the same expected value i don't understand what this means? can you explain ?

 

[Ray Vickson]

N, is N = sum{ I_j : j = 1...5}, so EN = sum{EI_j : j = 1...5}, and each I_j has the same expected value i don't understand what this means? can you explain ?

I_j is 1 if the jth part drawn is defective, and I_j = 0 otherwise. The sum of the I_j values is just the total number (N) of defectives drawn, and the expected value of the sum is the expected number (EN) of defectives drawn, which the OP wants to calculate. There is a fundamental property: the expected value of a sum is the sum of the expected values, so EN is the sum of the EI_j values. Finally, each I_j has the same probability distribution, so has the same expected value; in fact, EI_j = Prob{I_j = 1}.

RGV