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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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:…
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A car insurance company has determined that 9% of all drivers were involved in a car accident last year. Among the 10 drivers living on one particular street, 3 were involved in a car accident last year. If 10 drivers are randomly selected, what is the probability of getting 3 or more who were involved in a car accident last year? O a. O b. O C. O d. 0.4435 0.9548 0.0541 0.0452

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