Problems 15-25: Questions: a.) Which is the Random Variable for each problem? b.

Question

Problems 15-25:
Questions: a.) Which is the Random Variable for each problem? b.) Which type of distribution applies on each problem? c.) Parameters. d.) What does the problem ask for?
* YOU DON’T HAVE TO CALCULATE ANYHING * 5. A card is drawn from a well-shuffled deck of $2 play?ng cards the result record and the card replaced. If the experiment is repeated ive times, what is (obraining two spades and one heart? The surface of a circular dart board has a small center circle called the buil's-eye the probability und0 pe-shaped regions numbered from I to 20. Each of the pie-shaped repons is turther diided into ihree parts such that a person throwing a dart that lands on a speciied number scores the value of the number. double the number. or umber. depending on which of the three parts the dart falls. If a person hits the buil-eye with probability 001. hits a double with probability 0.10. hits a tripie with bility 0.05. and misses the dart board with probability002 what i he pra that seven throws will result in no buil's-eyes, no triples, a double twice and a compliete miss once? 17. According to the theory of genetics, a certain cross of guinea pigs will result in red black, und w hite ofspring in the ratio 3:4:4. Find the probability that among eight offspriang hve will be red. two black. and one white. 13. To avoid deiection ai customs, a traveler has piaced six narcotic tablets in a botthe containing nine vitamin pills that are similar in appearance. If the customs official selects three of the tablets at random for analysis, what is the probability that the traveler will be arrested for illegal possession of narcoties? 19. A company is interested in evaluating its current inspection procedure on shipments of 50 dentical items. The procedure is to take a sample of 5 and pass the shipment if no more than 2 are found to be defective. What proportion of 20. defective shipments will be accepied? 20. A manufacturing company uses an acceptance scheme on production items before they are shipped. The plan is a two-stage one. Boxes of 25 are readied for shipment and a sampe of 3 are tested for defectives. If any defectives are found, the entire box s sent back for 100*, screening. If no defectives are found, the box is shipped a) What is the probability that a box containing three defectives will be shipped? ibi What is the probability that a box containing only one defective will be sent back for screening 21. Suppose that the manufacturing company in Exercise 20 decided to change its accept ance scheme. Under the new scheme an inspector takes one at random inspects it and then replaces it in the box: a second inspector does likewise. Finally.a third inspector goes through the same procedure. The box is not shipped if any of the three a defective. Answer parts lal and ibi of Exercise 20 under this new plan. homeowner plants six buibs seiected at random from a box containing five buibs and four daffodil buibs. What is the probabiity that he planted two dañodd buibs and four tulip buibs? Find the probability of obtaining 2 ones. I two. I three. 2 fours 3 ives, and I siz in 10 rolls of a balanced die? ormula for the probability distribution of the random variabie X representing the number of doctors on the committee. Find PA2SX s 3 4. A random committee of size 3 is selected from four doctors and two nurses Wrnte a From a lot of 10 missiles. 4 are selected at random and fired. If the lor contains 3 defective missiles that will not fire. what is the probability that a) All 4 wil fire? ib? At most 2 will not fire

Explanation / Answer

25) Let X be the random variable that number of defective missilies that will not fire.

Here X ~ Binomial (n = 10, p(fire) = 4/10)

This is the discrete random variable.

P(not fire) = 6/10

To determine probability.

24) Let ?? be the number of doctors on the committee. ?? is a hypergeometric variable with

?? ? ????????????????????????????(?? = 3, ?? = 6,?? = 4)

This is the discrete random variable.

The pmf of X=x is,

P(X = x) = (k C x) * [(N-k) C (n-x)] / (N C n)

23) X = (X1, X2,X3,X4,X5,X6)

Here X follows multinomial distribution.

Here Xk is the amount of k's we obtained.

17) total number of offspring = 8+4+4 = 16

P(red) = 8/16 = 1/2
P(black) = 4/16 = 1/4

P(white) = 4/16 = 1/4

Each birth an Independent Event(with a probability of its own)

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