2 questions
Roll a fair 6 sided die 3 itmes. What is the probablity that you obtain at least one 5? Use the binomial.
A 28 yr old man pays $208 for a 1 year life insurance policy with coverage of $110,000. If the probability that he will live through the year is .9993, what is the expected value of the policy?
207.85
77
131
109.923
667.32
if its 364.25 days in a year then each day would be 301.99, and .9993 of 1 year would be 363.99
so 301.99 x 363.99 = 109,921.35
so im guessing the 109.923 would be the correcdt answer... unless im going about solving this problem wrong...
1) The probability of NOT getting a five on a single roll is 5/6. The probability of NOT getting a five on 3 rolls is (5/6)^3 = .5787. Sooo, the probability of getting at least 1 five is 1-.5787 = 4213.
2) The probability of dying is .0007 The payoff from dying is 110,000. Sooo, the expected value of the policy, regardless of the price paid, is .0007*110000 = 77.
(n+1)!/(n-2)!
For the first question, to calculate the probability of getting at least one 5 when rolling a fair 6-sided die 3 times, we can use the binomial distribution.
The formula for the probability of getting exactly k success (in this case, rolling a 5) in n trials is:
P(k) = C(n, k) * p^k * (1-p)^(n-k)
Where C(n, k) is the combination of n objects taken k at a time, p is the probability of success on a single trial, and (1-p) is the probability of failure on a single trial.
In this case, the probability of rolling a 5 on a fair 6-sided die is 1/6, so p = 1/6. We want to calculate the probability of getting at least one 5, so we need to find P(k>=1).
To calculate this, we can find the probability of getting 0 5's and subtract it from 1.
P(k>=1) = 1 - P(k=0)
P(k=0) = C(3, 0) * (1/6)^0 * (5/6)^(3-0)
C(3, 0) = 1, as there is only one way to choose 0 successes from 3 trials.
Plugging in the values:
P(k=0) = 1 * (1/6)^0 * (5/6)^3 = (5/6)^3 = 125/216 ≈ 0.5787
P(k>=1) = 1 - P(k=0) = 1 - 125/216 ≈ 0.4213
So, the probability of obtaining at least one 5 when rolling a fair 6-sided die 3 times is approximately 0.4213.
For the second question, you correctly identified the probability of living through the year as 0.9993. The expected value of the policy can be calculated as the product of the probability of each outcome (living or dying) and the payoff associated with that outcome.
The probability of living is 0.9993, and the payoff from living is $0. Therefore, the contribution to the expected value from living is 0.9993 * 0 = $0.
The probability of dying is 1 - 0.9993 = 0.0007, and the payoff from dying is $110,000. Therefore, the contribution to the expected value from dying is 0.0007 * $110,000 = $77.
The total expected value of the policy is the sum of the contributions from living and dying:
Expected value = $0 + $77 = $77.
So, the expected value of the policy is $77.