In: Statistics and Probability
A weapons manufacturer uses a liquid propellant that can get mixed with another liquid to produce a contaminated cartridge. A statistician found that 24% of the cartridges in the particular lot were contaminated. Suppose you randomly sample (without replacement) gun cartridges from this lot until you find a contaminated one. Let x be the number of cartridges sampled until a contaminated one is found. It is known that the probability distribution for x is given by the formula shown below. Complete parts a through c. p left parenthesis x right parenthesis equals left parenthesis 0.24 right parenthesis left parenthesis 0.76 right parenthesis Superscript x minus 1, xequals1, 2 ,3 ,... a. Find p(1). Interpret this result. p(1)equals nothing (Round to three decimal places as needed.) What is the correct interpretation for p(1)? A. This value is the probability that one would encounter a contaminated cartridge on the first trial. B. This value is the probability that one would encounter a contaminated cartridge in one hundred trials. C. This value is the probability that one would encounter a non-contaminated cartridge on the first trial. b. Find p(5). Interpret this result. p(5)equals nothing (Round to three decimal places as needed.) What is the correct interpretation for p(5)? A. This value is the probability that one would first encounter 5 contaminated cartridge in one hundred trials. B. This value is the probability that one would first encounter a contaminated cartridge on the fifth trial. C. This value is the probability that one would first encounter a non-contaminated cartridge on the fifth trial. c. Find P(xgreater than or equals2). Interpret this result. Upper P left parenthesis x greater than or equals 2 right parenthesisequals nothing (Round to three decimal places as needed.) What is the correct interpretation for P(xgreater than or equals2)?
Let x be the number of cartridges sampled until a contaminated one is found.
It is known that the probability distribution for x is given by the formula shown below.
a. Find p(1).
We get p(1)= 0.240
According to the event X, defined in the problem,
A. This value is the probability that one would encounter a contaminated cartridge on the first trial.
b.
We get p(5)= 0.080
B. This value is the probability that one would first encounter a contaminated cartridge on the fifth trial.
c. Find
= 1 - 0.24
= 0.76
= 0.760
Interpretation:
C. This value is the probability that one would first encounter a contaminated cartridge on 2nd or later trial, which is equal to the complementary of the probability that one would encounter a contaminated cartridge on the 1st trial itself.