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- When the health department tested private wells in a county for two impurities commonly found in drinking water, it found that 10% of the wells had neither impurity, 90% had impurity A, and 20% had impurity B. (Obviously, some had both impurities.) If a well is randomly chosen from those in the county, find the probability distribution for Y, the number of impurities found in the well. P(no impurities)= P(exactly one impurity)= P(both impurities)=O An oll company purchased an option on land in Alaska, Preliminary geologic studies assigned the following prior probabilities. P(high-quality oil) = 0.45 P(medium-quality oil) = 0.20 P(no oil) = 0.35 a. What is the probability of finding oll (to 2 decimals)? .65 b. After 200 feet of drilling on the first well, a soil test is taken, The probabilities of finding the particular type of soil identified by.the test are given below. P(soil|high-quality oil) = 0.25 %! P(soil medium-quality oil) = 0.85 P(soil no oil) = 0.25 Given the soil found in the test, use Bayes' theorem to compute the following revised probabilities (to 4 decimais). P(high-quality oiljsoil) P(medium-quality oil soil) P(no oilsoil) What is the new probability of finding oil (to 4 decimals)? According to the revised probabilities, what is the quality of oil that is most likely to be found? Select your answer O teon KeyA biologist estimates that the chance of germination for a type of bean seed is 0.7. A student was given 6 seeds. Let X be the number of seeds germinated from 6 seeds. Assuming that the germination of seeds are independent, explain why the distribution of X is binomial. What are the values of n and p? What are the probabilities that he gets (a) all seeds germinated, (b) just one seed not germinated, and (c) at most four seeds germinated?
- Yes, because (0.63)(0.63) = 0.497 No, because (0.63)(0.63) ≠ 0.497 Yes, because 63% + 63% = 49.7% No, because 63% + 63% ≠ 49.7%Q.No.3. (a) A bag contains 12 balls, 4 of which are red, 3 black and 5 green. Three balls are drawn from the bag. Find the probability that (i) all three are green (ii) The first is red, second is green and third is black. (b) A sale person has a 40% chance of making a sale to any customer who is called upon. If 18 calls are made, what is the probability that (i) at most 3 sales are made (ii) at least 17 sales are made.An experiment can result in one of five equally likely simple events, E₁, E₂, A: E₂, E4 2' P(A) = = 0.4 B: E₁, E3, E4, E5 P(B) = 0.8 C: E₂, E3 Refer to the following probabilities. P(A n B) = 0.2 P(AIB) = 0.25 P(BIA) P(BU C) = 1 P(BIC) 0.5 P(CIB) 0.25 Use the Addition and Multiplication Rules to find the following probabilities. (a) P(AUB) (b) P(An B) = 0.5 P(A U B) = ---Select--- + ---Select--- ✓ P(A n B) = ---Select--- (c) P(B n C) Yes P(B n C) = ---Select--- No P(C) = 0.4 . P(B) = = P(C) = = ---Select--- V = Do the results agree with those obtained by listing the simple events in each? P(A U B) = P({E₁, E₂, E3, E4, E5}) = 1 P(A n B) = P({E₁}) = 0.2 P(B n C) = P({E3}) = 0.2 .., E5. Events A, B, and C are defined as follows.
- Exercise 4. A certain species of plant always has either three or five leaves. The number is random, with P(3 leaves) 0.4 and P(5 leaves) = 0.6. Each plant has a flower which, randomly, is either open or closed, with probabilities P(open) = 0.8 and P(closed) = 0.2. A botanist collects 1000 randomly chosen plants from this species and nds the following distribution of traits: open closed N3,closed 3 leaves N3,open 5 leaves NE,open N5, closed a) Assuming the two traits are independent, determine the expectations of the counts №3,open, N3,closed, N5,0-en, and N5,close in the table. b) Determine an approximate value for the probability P(N3,open > 340). Exercise 5. Assume that we have observed the following values from a normal distribution with known variance o2 = and unknown mean . 1.23 -0.67 1.16 1.67 0.24 2.99 0.02 .17 0.27 21. Test the hypothesis Ho: = 0 against the alternative H₁: #0 at significance level a = 5%. Exercise 6. Let 0> 0 and XU[0,0], i.e. X is uniformly distributed on…Let A, B and C be events from a common sample space such that: P(A) = 0.72, P(B) = 0.56, P(C) =0.25, P(ANB) = 0.42. P(ANC) = 0.18, P(BOC) =0 %3D Compute P(AUBUC).A hospital reports that two patients have been admitted who have contracted Crohn's disease. Suppose our experiment consists of observing whether each patient survives or dies as a result of the disease. The simple events and probabilities of their occurrences are shown in the table (where S in the first position means that patient 1 survives, D in the first position means that patient 1 dies, etc.). Simple Events Probabilities SS 0.52 SD 0.20 DS 0.14 DD 0.14 Find the probability that both patients survive.
