Suppose a cashier gives you a dollar coin in change. Because you have no reason todistrust this coin your prior is very high that this is a fair coin.In fact, your prior belief of the probability of the coin to give heads p(H) is thefollowing probability distribution:p (p(H) = 0) = C 050 (1-0)5⁰ dewhere C is just a normalization constant for the distribution to sum to 1:61/11211/1259/108You decide to flip the coin 10 times, and all 10 times it turns out head. What is yourmaximum a posteriori estimate (MAP) for the probability of the coin to give heads?160/110C =01/21S²050 (1-0) 50

To find the Maximum A Posteriori (MAP) estimate for the probability of the coin to give heads (D(H))…

Answered: se you ave no reaso Suppose a cashier…

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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I need help with this. Answer is NOT 0.0080
Complete the following probability distribution. ENTER EACH OF YOUR ANSWER TO 2 DECIMAL PLACE
note: the current answers are wrong
Suppose you are given the following facts. For a given disease, there is a test which is 80% reliable (Pr(Y|D) = .8). How high does the base rate for the disease need to be for the probability of a false positive to be 20%?
The probability that a randomly selected teenager watched a rented video at least once during a week was 0.69. What is the probability that at least 7 teenagers in a group of 9 watched a rented movie at least once last week? (Round your answer to four decimal places.) P(X ≥ 7) =
An unfair die looks like an ordinary six-sided die but the outcomes are not equally likely. The probability distribution of the face value, XX, is as follows: xixi 1 2 3 4 5 6 Total P(X=xi)P(X=xi) 0.15 0.16 0.33 0.2 0.11 0.05 1 The expected value is computed, E[X]=3.11E[X]=3.11. To find the variance and standard deviation the following table was set up: xixi P(X=xi)P(X=xi) (xi−E[X])2(xi-E[X])2 (xi−E[X])2P(X=xi)(xi-E[X])2P(X=xi) 1 0.15 (1−3.11)2=(−2.11)2=4.4521(1-3.11)2=(-2.11)2=4.4521 4.4521⋅0.15=0.66784.4521⋅0.15=0.6678 2 (2−3.11)2=(−1.11)2=1.2321(2-3.11)2=(-1.11)2=1.2321 1.2321⋅0.16=0.19711.2321⋅0.16=0.1971 3 0.33 (3−3.11)2=(−0.11)2=(3-3.11)2=(-0.11)2= 0.0121⋅0.33=0.0040.0121⋅0.33=0.004 4 0.2 (4−3.11)2=(0.89)2=0.7921(4-3.11)2=(0.89)2=0.7921 0.7921⋅0.2=0.7921⋅0.2= 5 (5−3.11)2=(1.89)2=3.5721(5-3.11)2=(1.89)2=3.5721 3.5721⋅0.11=0.39293.5721⋅0.11=0.3929 6 0.05 8.3521⋅0.05=0.41768.3521⋅0.05=0.4176 E[X]=3.11E[X]=3.11 Total:…
I need help with parts b to d please, I posted this twice and got wrong answers. Answer for part a) is 1/7
Andrew finds that on his way to work his wait time for the bus is roughly uniformly distributed between 6 minutes and 9 minutes. One day he times his wait and writes down the number of minutes ignoring the seconds. Round solutions to three decimal places, if necessary. 0.3 0.25 0.2- 0.15 0.1- 0.05 6. 7 8. Wait time measured in minutes rounded down What is the probability that Andrew waits for 9 minutes? P(X = 9) = What is the probability that Andrew waits between 6 and 8 minutes? P(6 < X < 8) = W

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