AS .801 Fall20. T Homework#3(1)-word Matt Robinson -- × Find pboard For Style a
ID: 2929579 • Letter: A
Question
AS .801 Fall20. T Homework#3(1)-word Matt Robinson -- × Find pboard For Style a Ed tinp I play a special chess tournament, where you keep playing until you lose a game, then you get knocked out. Since draws are possible, theoretically I could play an infinite number of games and never get knocked out. There is a 90% chance I don't lose any given individual game. a. What is the probability I lose my 11th game, b. Given I lose a game # divisible by 3 (ie. I lose game number 3, or 6, or 9 etc.), 2. what is the probability I lose game #6? E O lype here to searchExplanation / Answer
Question 2
Pr(I don't loose any individual game) = 0.90
(a) Pr(I will lose 11th game) = Pr(won all initial 10 games) * Pr( lost 11th game) = 0.9010 * 0.10 = 0.0349
(b) Given that i loose a game divisble by 3.
Pr(lossing game number 3) = Pr( X = 3) + Pr( X = 6) + Pr( X=9) + .......
= (0.9)2 0.10 + (0.9)5 0.10 + (0.9)8 (0.10) + ....
= (0.9)2 0.10 [ 1 + 0.93 + 0.96 + ...]
= (0.9)2 * 0.10 * [1/ (1 - 0.93)]
= 0.2989
Pr(X = 3) = (0.9)2 * 0.10 = 0.081
Pr( X = 3 l X = 3k) = (0.9)2 0.10 / [(0.9)2 * 0.10 * 1/ (1 - 0.93)] = 0.271