Question

In: Statistics and Probability

Sample size (n) Sample mean (x̄) Sample Variance () Stun gun 10 13.6667 3.5152 Poisoned M&M...

Sample size (n)

Sample mean (x̄)

Sample Variance ()

Stun gun

10

13.6667

3.5152

Poisoned M&M

10

12.25

3.8409

Murder hornet

10

14.0833

0.5184

(1) If the high voltage stun gun was the murder weapon, then victim must have been electrocuted. Suppose that it take 5 direct shocks with the stun gun to kill a victim and that the probability of a direct shock on any given attempt is 0.8.

Assume that the murderer took 6 attempts.

(a) Which probability distribution should you use?

(b) What is the probability of success? How many trials?

(c)If the murderer took 6 attempts, calculate the probability of killing the victim (that is, the probability of obtaining at least 5 direct shocks).

(d) If the poisoned M&M was the murder weapon, it is likely that the suspect created several poison M&Ms to increase the odds that the victim would be poisoned. From the M&Ms that were found at the scene, the police determined that the ratio of M&Ms that were poisoned to not poisoned was 4 to 1. Use this ratio to determine the probability of choosing a poisoned M&M.

(e) If the murder hornet was the murder weapon, then the suspect must have trapped the victim in a room with the agitated hornet.

P(being severely allergic to insect stings) = 0.001

P(dying from murder hornet sting AND being severely allergic to insect stings) = 0.0009

After obtaining the victim’s medical records, the police determine that he is severely allergic to insect stings. Given that he is severely allergic to insect stings, what is the probability of the victim from a murder hornet sting?

P(dying from a murder hornet sting | being severely allergic to insect stings) =

Solutions

Expert Solution

1) Given  the high voltage stun gun was the murder weapon, then victim must have been electrocuted. Suppose that it take 5 direct shocks with the stun gun to kill a victim and that the probability of a direct shock on any given attempt is 0.8.

Assume that the murderer took 6 attempts.

(a) Here, we have to use the binomial probability distribution by definining the random variable X as number of successes means that number of direct shocks to kill the victim by sten gun out of n trials with p as probability of succes at each trial then,

P(X=r) =nCr p^r (1-p)^(n-r), r=0,1,2,...,n

(b) The probability of success is the probability of a direct shock on any given attempt is p=0.8.

The number of trials = n=6 attempts

(c) If the murderer took 6 attempts, then the probability of killing the victim (that is, the probability of obtaining at least 5 direct shocks) is

P(X>=5) = P(X=5) + P(X=6)

=6C5 0.8^5 (1-0.8)^(6-5)+6C6 0.8^6 (1-0.8)^(6-6)

=6* 0.8^5*0.2+1*0.8^6*1

=0.6553

(d) It is given that from the M&Ms that were found at the scene, the police determined that the ratio of M&Ms that were poisoned to not poisoned was 4 to 1.

Then the ratio to determine the probability of choosing a poisoned M&M is

4/(4+1) = 4/5 = 0.8

(e) If the murder hornet was the murder weapon, then the suspect must have trapped the victim in a room with the agitated hornet, given that

P(being severely allergic to insect stings) = 0.001

P(dying from murder hornet sting AND being severely allergic to insect stings) = 0.0009

After obtaining the victim’s medical records, the police determine that he is severely allergic to insect stings. Given that he is severely allergic to insect stings, then the probability of the victim from a murder hornet sting is

P(dying from a murder hornet sting | being severely allergic to insect stings) =

P(dying from a murder hornet sting AND being severely allergic to insect stings) / P(being severely allergic to insect stings) =

= 0.0009/0.001=0.9


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