A p a a ω n k 2 n problem 2 suppose we flip a coin n

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A: P ( A ) = | A | / | Ω | = ( n k ) / 2 n . Problem 2 Suppose we flip a coin n times independently. The probability of getting a head is p in each flip. Let X be the number of heads. What is P ( X = k ) for k = 0 , 1 , ..., n ? A: P ( X = k ) = ( n k ) p k (1 - p ) n - k . The reason is that the number of sequences with k heads is ( n k ) . The probability of getting each sequence with k heads is p k (1 - p ) n - k . Problem 3 Suppose we have a population of R red balls and B blue balls, and N = R + B . Suppose we randomly sample n balls without replacement. (1) Let X be the number of red balls. What is P ( X = k ), for k = 0 , 1 , ..., n ? You can assume that both R and B are larger than n . A: The sample space Ω consists of all combinations of n balls chosen from the N balls in the population. The event A = { X = k } consists of all the combinations with k red balls. P ( X = k ) = | A | / | Ω | = ( R k )( B n - k ) / ( N n ) . (2) Fix p = R/N , and let N goes to infinity, then what is the limit of P ( X = k )? Compare the result with that of Problem 2. A: R = Np . As N → ∞ , R also goes to infinity. P ( X = k ) = R ( R - 1) ... ( R - k +1) k ! B ( B - 1) ... ( B - ( n - k )+1) ( n - k )! N ( N - 1) ... ( N - n +1) n ! = n ! k !( n - k )! R N R - 1 N - 1 ... R - k + 1 N - k + 1 B N - k B - 1 N - k - 1 ... B - ( n - k ) + 1 N - n + 1 = n ! k !( n - k )! p k (1 - p ) n - k 3
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STATS 100A HW3 Problem 1 Prove (1) P ( A c | B ) = 1 - P ( A | B ). A: P ( A c | B ) + p ( A | B ) = P ( A c B ) /P ( B ) + P ( A B ) /P ( B ) = P ( B ) /P ( B ) = 1. (2) P ( A B | C ) = P ( A | C ) + P ( B | C ) - P ( A B | C ). A: P ( A B | C ) = P (( A B ) C ) /P ( C ) = P (( A C ) ( B C )) /P ( C ) = ( P ( A C ) + P ( B C ) - P ( A B C )) /P ( C ) = P ( A | C ) + P ( B | C ) - P ( A B | C ). (3) P ( A B C ) = P ( A ) P ( B | A ) P ( C | A B ). A: P ( A ) P ( B | A ) P ( C | A B ) = P ( A )[ P ( A B ) /P ( A )][ P ( A B C ) /P ( A B )] = P ( A B C ). Problem 2 Suppose an urn has r red balls and b black balls. We randomly pick a ball, and then put two balls of the same color back into the urn. We repeat this process. (1) What is the probability that the second pick is red? A: Let R 1 be the event that the first pick is red, and B 1 be the event that the first pick is black. Similarly let R 2 be the event that the second pick is red, and B 2 be the event that the second pick is black, and so on. P ( R 2) = P ( R 1) P ( R 2 | R 1) + P ( B 1) P ( R 2 | B 1) = r r + b r + 1 r + b + 1 + b r + b r r + b + 1 = r r + b (2) What is the probability that the third pick is red? P ( R 3) = P ( R 1) P ( R 2 | R 1) P ( R 3 | R 1 , R 2) + P ( R 1) P ( B 2 | R 1) P ( R 3 | R 1 , B 2) + P ( B 1) P ( R 2 | B 1) P ( R 3 | B 1 , R 2) + P ( B 1) P ( B 2 | B 1) P ( R 3 | B 1 , B 2) = r r + b r + 1 r + b + 1 r + 2 r + b + 2 + r r + b b r + b + 1 r + 1 r + b + 2 + b r + b r r + b + 1 r + 1 r + b + 2 + b r + b b + 1 r + b + 1 r r + b + 2 = r r + b Problem 3 Suppose we perform a random walk over three states Ω = { 1 , 2 , 3 } . Let X t be the state at step t . Suppose we start from 1, i.e., X 0 = 1. Suppose at each step, we randomly move to one of the other two states with equal probability, regardless of the past history. That is, P ( X t +1 = y | X t = x, X t - 1 , X t - 2 , ..., X 0 ) = P ( X t +1 = y | X t = x ). Let K ( x, y ) = P ( X t +1 = y | X t = x ), where x, y Ω. (1) Write down K as a 3 × 3 matrix. A: K = 0 1 / 2 1 / 2 1 / 2 0 1 / 2 1 / 2 1 / 2 0 (2) Prove P ( X t +1 = y ) = x Ω P ( X t = x ) K ( x, y ) for any y Ω.
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  • Fall '07
  • Wu
  • Probability, Notes, Probability theory, WU, var

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