Please explain how you got the answer step by step Random words are created by c
ID: 3207722 • Letter: P
Question
Please explain how you got the answer step by step
Random words are created by choosing the first letter using the following table of single letter statistics then each subsequent letter is chosen using the table of biletter statistics shown to the right. How much information do you learn on average when you are told a two letter word given that you know that the word begins with the letter A? How much information do you learn on average when you are told a two letter word given that you know that the word ends with the letter D? How much information do you learn on average when you are told a two letter word given that you know at least one of the letters is a D?Explanation / Answer
PART (a) -If word begins with letter A then second letter can either be A,B,C,D. Second letter will be chosen from biletter statistics.Second letter will be A with probability of 4/10. Next letter can't be letter B because as it can be seen in statistics given in question, when A comes first B can't follow it. Next letter will C with probability of 1/10 and it'll be letter D with a probability of 5/10. SO it can be summarised that on an average two letter word will be as follows , with their given chances:
AA- .4
AB- 0
AC- .1
AD- .5
PART (b)- If two letter word ends with letter D, then only first letter can be either A or B.If first letter is A then word will be AD with probability equal to =(4/10)*(5/10) =1/5.If first letter is B then word will be BD with probability of (1/10)*(6/10)=6/100. Using BAYE'S THEOREM , we get :
if word ends with D , then word is AD with probability =
(1/5)/(1/5+6/100) =10/13
if word ends with D , then word is BD with probability=
(6/100)/(1/5+6/100) = 3/13.
PART (C) - Word containing atleast one letter D possible are - AD, BD, DA and DC. Probability that two letter word has atleast one letter D is = (4/10)*(5/10) + (1/10)*(6/10) + (3/10)*(1) = 56/100.
Since we know that atleast one letter is D , so USING AGAIN BAYE'S THEOREM, we obtain,
probability that word will be AD given that there is atleast one letter D is=
(4/10)*(5/10)/(56/100) = 20/56
probability that word will be BD given that there is atleast one letter D is =
(1/10)*(6/10)/(56/100) = 6/56
probability that word will be DA given that there is atleast one letter D is=
(3/10)*(1/10)/(56/100) = 3/56
probability that word will be DC given that there is atleast one letter D is=
(3/10)*(9/10)/(56/100) = 27/56
***FINISHED***