In: Statistics and Probability
QUESTION THREE
At least one head
Question (3A)
The probability of you experiencing COVID-19 is 0.5% which is 0.005
So the probability of you not experiencing COVID-19 = 1- 0.005 = 0.995
Now we calculate the expected value of the insurance for you
The value for the You column is (Benefits - Cost)
If you experience COVID-19, then the effective value for you = 300000 - 1500 = 298500
You are getting a beneft of 300000 and you are paying 1500, so the effective value for you = 300000 - 1500 = 298500
If you do not experience COVID-19, then the effective value for you = 0 - 1500 = -1500
You won't get anything if you don't experience COVID-19, so benefit is 0, cost is 1500
Case | Probability P(x) | Effective Value for you (V) | P(x) * E |
Experiencing COVID-19 | 0.005 | 298500 | 1492.5 |
Not Experiencing COVID-19 | 0.995 | -1500 | - 1492.5 |
P(x) * E = 0 |
The expected value from the insurance for you = P(x) * E = 0
So the insurance is fair since you are getting insured and you would be risk-free
Question (3B)
Given that a coin is biased so that it has a 60% chance of landing on heads
So P(H) = 0.6
The coin toss is a binomial distribution since there are only 2 outcomes possible here. Head and Tail
Probability of x successes in n trails for a binomial distribution is * px * qn-x
= n! / x! (n-x)!
Here p is probability of success and q is 1-p
Question (a)
Probability of three heads
Here p = 0.6 probability of a head
q = 0.4
n = 3 number of times a coin is tossed
x = 3 number of heads or number of successess
So Probability of three heads = * 0.63 * 0.43-3
= 3! / 3! (3-3)! * 0.216 * 1
= 0.216
So Probability of three heads = 0.216
Question (b)
Probability of two heads and a tail is same as calculating the probability of two heads since the third one would automatically be a trial
Here p = 0.6 probability of a head
q = 0.4
n = 3 number of times a coin is tossed
x = 3 number of heads or number of successess
So Probability of two heads = * 0.62 * 0.43-2
= 3! / 2! (3-2)! * 0.36 * 0.4
= 0.432
So Probability of two heads and one tail = 0.432
Question (c)
Probability of atleast 1 head is obtained by adding porbabilty of 1 head and 2 head and 3 heads
Here p = 0.6 probability of a head
q = 0.4
n = 3 number of times a coin is tossed
x = 3 number of heads or number of successess
So Probability of atleast 1 head = * 0.61 * 0.43-1 + * 0.62 * 0.43-2 + * 0.63 * 0.43-3
= 3! / 1! (3-1)! * 0.6 * 0.16 + 3! / 2! (3-2)! * 0.36 * 0.4 + 3! / 3! (3-3)! * 0.216 * 1
= 0.288 + 0.432 + 0.216
= 0.936
So Probability of atleast 1 head = 0.936