MATLAB: An Introduction with Applications
6th Edition
ISBN: 9781119256830
Author: Amos Gilat
Publisher: John Wiley & Sons Inc
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- Results from trials for a new pill for sleeping are given in this table. slept better (B) no improvement (N) sleeping pill (S) 144 21 sugar tablet (T) 55 89 What is the probability that the subjects slept better if they took the sleeping pill = P(B|S) ? What is the probability that the subjects took the tablet if they slept better = P(T|B)? Give your answer as a fraction.arrow_forwardUse the Test for Independence to determine if events A and B are independent. P(A) = 0.7, P(B) = 0.8, P(An B) = 0.56 Select the correct choice below and fill in the answer box(es) to complete your choice. (Type an integer or a decimal.) O A. Events A and B are not independent because P(A U B)= | and P(A) + P(B) - P(An B)= O B. Events A and B are independent because P(An B) = 0.56 and P(A).P(B) = O C. Events A and B are independent because P(A U B)= and P(A) + P(B)- P(An B) = O D. Events A and B are not independent because P(An B) = 0.56 and P(A) P(B) =.arrow_forwardSuppose we know that the probabilities P(A) = .5, P(B|A) = .3 and P(A or B) =.3. What is the value for P(A and B)? What is the value for P(B)?arrow_forward
- T1.8 Let A and B be two events with probabilities P(A)=.35, P(B)=.40 and P(A U B)=.6. a) Find P(A intersection B) b) Find P(A | B) c) Are events A and B independent or not?arrow_forwardK Find P(A or B or C) for the given probabilities. P(A)=0.39, P(B) = 0.22, P(C) = 0.11 P(A and B) = 0.12, P(A and C) = 0.02, P(B and C) = 0.07 P(A and B and C) = 0.01 P(A or B or C) =arrow_forward9. By rewriting the formula for the Multiplication Rule, you can write a formula for finding P(A and B) conditional probabilities. The conditional probability of event B occurring, given that event A has occurred, is P(B| A) = P(A) the information below to find the probability that a flight arrives on time given that it departed on time. The probability that an airplane flight departs on time is 0.92. The probability that a flight arrives on time is 0.86. The probability that a flight departs and arrives on time is 0.83. The probability that a flight arrives on time given that it departed on time is (Round to the nearest thousandth as needed.) . Usearrow_forward
- Assume that a researcher randomly selects 14 newborn babies and counts the number of girls selected, x. The probabilities corresponding to the 14 possible values of x are summarized in the given table. Answer the question using the table. Probabilities of Girls x(girls) P(x) x(girls)| P(x) |x(girls)| P(x) 0.000 0.122 10 0.061 1 0.001 0.183 11 0.022 2 0.006 7 0.210 12 0.006 3 0.022 8 0.183 13 0.001 4 0.061 9 0.122 14 0.000 Find the probability of selecting 12 or more girls. 0.001 O 0.022 0.006 O 0.007arrow_forwardq15A -arrow_forwardSuppose that events A and B are mutually exclusive with P(A) = 1 2 and P(B) = 1 6 . (Enter your answers as fractions.) (a) Are A and B independent events? Explain how you know. Since the events are mutually exclusive we know P(A|B) = which P(A) and thus the events are . (b) Are A and B complementary events? Explain how you know. We know that P(A) + P(B) = . And, since P(A) + P(B) 1, the events A and B complementary events.arrow_forward
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