bfn X - Technical University of Denmark

IMM - DTU Question a) X

02405 Probability 2004-3-3 BFN/bfn

IMM - DTU Question a) X the number of heads in 9 tosses is

02405 Probability 2004-3-3 BFN/bfn

IMM - DTU

02405 Probability 2004-3-3 BFN/bfn

Question a) X the number of heads in 9 tosses is binominally distributed, thus

02405 Probability 2004-3-3 BFN/bfn

IMM - DTU

Question a) X the number of heads in 9 tosses is binominally distributed, thus P (X = 5) =

IMM - DTU

02405 Probability 2004-3-3 BFN/bfn

Question a) X the number of heads in 9 tosses is binominally distributed, thus   9 P (X = 5) = 5

IMM - DTU

02405 Probability 2004-3-3 BFN/bfn

Question a) X the number of heads in 9 tosses is binominally distributed, thus   9 5 P (X = 5) = p 5

IMM - DTU

02405 Probability 2004-3-3 BFN/bfn

Question a) X the number of heads in 9 tosses is binominally distributed, thus   9 5 P (X = 5) = p (1 − p)4 5

IMM - DTU

02405 Probability 2004-3-3 BFN/bfn

Question a) X the number of heads in 9 tosses is binominally distributed, thus   9 5 P (X = 5) = p (1 − p)4 5 Question b) Y the number of tosses for the first head is

02405 Probability 2004-3-3 BFN/bfn

IMM - DTU

Question a) X the number of heads in 9 tosses is binominally distributed, thus   9 5 P (X = 5) = p (1 − p)4 5 Question b) Y the number of tosses for the first head is geometrically distributed, thus P (Y = 7) =

IMM - DTU

02405 Probability 2004-3-3 BFN/bfn

Question a) X the number of heads in 9 tosses is binominally distributed, thus   9 5 P (X = 5) = p (1 − p)4 5 Question b) Y the number of tosses for the first head is geometrically distributed, thus P (Y = 7) = (1 − p)

IMM - DTU

02405 Probability 2004-3-3 BFN/bfn

Question a) X the number of heads in 9 tosses is binominally distributed, thus   9 5 P (X = 5) = p (1 − p)4 5 Question b) Y the number of tosses for the first head is geometrically distributed, thus P (Y = 7) = (1 − p)6

IMM - DTU

02405 Probability 2004-3-3 BFN/bfn

Question a) X the number of heads in 9 tosses is binominally distributed, thus   9 5 P (X = 5) = p (1 − p)4 5 Question b) Y the number of tosses for the first head is geometrically distributed, thus P (Y = 7) = (1 − p)6 p

02405 Probability 2004-3-3 BFN/bfn

IMM - DTU

Question a) X the number of heads in 9 tosses is binominally distributed, thus   9 5 P (X = 5) = p (1 − p)4 5 Question b) Y the number of tosses for the first head is geometrically distributed, thus P (Y = 7) = (1 − p)6 p

Question c) Z the number of tosses to get 5 heads follows a

02405 Probability 2004-3-3 BFN/bfn

IMM - DTU

Question a) X the number of heads in 9 tosses is binominally distributed, thus   9 5 P (X = 5) = p (1 − p)4 5 Question b) Y the number of tosses for the first head is geometrically distributed, thus P (Y = 7) = (1 − p)6 p

Question c) Z the number of tosses to get 5 heads follows a negative binomial distribution P (Z = 12) =

IMM - DTU

02405 Probability 2004-3-3 BFN/bfn

Question a) X the number of heads in 9 tosses is binominally distributed, thus   9 5 P (X = 5) = p (1 − p)4 5 Question b) Y the number of tosses for the first head is geometrically distributed, thus P (Y = 7) = (1 − p)6 p

Question c) Z the number of tosses to get 5 heads follows a negative binomial distribution  11 P (Z = 12) =

IMM - DTU

02405 Probability 2004-3-3 BFN/bfn

Question a) X the number of heads in 9 tosses is binominally distributed, thus   9 5 P (X = 5) = p (1 − p)4 5 Question b) Y the number of tosses for the first head is geometrically distributed, thus P (Y = 7) = (1 − p)6 p

Question c) Z the number of tosses to get 5 heads follows a negative binomial distribution   11 P (Z = 12) = 4

IMM - DTU

02405 Probability 2004-3-3 BFN/bfn

Question a) X the number of heads in 9 tosses is binominally distributed, thus   9 5 P (X = 5) = p (1 − p)4 5 Question b) Y the number of tosses for the first head is geometrically distributed, thus P (Y = 7) = (1 − p)6 p

Question c) Z the number of tosses to get 5 heads follows a negative binomial distribution   11 P (Z = 12) = (1 − p) 4

IMM - DTU

02405 Probability 2004-3-3 BFN/bfn

Question a) X the number of heads in 9 tosses is binominally distributed, thus   9 5 P (X = 5) = p (1 − p)4 5 Question b) Y the number of tosses for the first head is geometrically distributed, thus P (Y = 7) = (1 − p)6 p

Question c) Z the number of tosses to get 5 heads follows a negative binomial distribution   11 P (Z = 12) = (1 − p)7 4

IMM - DTU

02405 Probability 2004-3-3 BFN/bfn

Question a) X the number of heads in 9 tosses is binominally distributed, thus   9 5 P (X = 5) = p (1 − p)4 5 Question b) Y the number of tosses for the first head is geometrically distributed, thus P (Y = 7) = (1 − p)6 p

Question c) Z the number of tosses to get 5 heads follows a negative binomial distribution   11 P (Z = 12) = (1 − p)7 p 4

IMM - DTU

02405 Probability 2004-3-3 BFN/bfn

Question a) X the number of heads in 9 tosses is binominally distributed, thus   9 5 P (X = 5) = p (1 − p)4 5 Question b) Y the number of tosses for the first head is geometrically distributed, thus P (Y = 7) = (1 − p)6 p

Question c) Z the number of tosses to get 5 heads follows a negative binomial distribution   11 P (Z = 12) = (1 − p)7 p5 4

IMM - DTU

02405 Probability 2004-3-3 BFN/bfn

Question a) X the number of heads in 9 tosses is binominally distributed, thus   9 5 P (X = 5) = p (1 − p)4 5 Question b) Y the number of tosses for the first head is geometrically distributed, thus P (Y = 7) = (1 − p)6 p

Question c) Z the number of tosses to get 5 heads follows a negative binomial distribution   11 P (Z = 12) = (1 − p)7 p5 4 Question d) X1 the number of

IMM - DTU

02405 Probability 2004-3-3 BFN/bfn

Question a) X the number of heads in 9 tosses is binominally distributed, thus   9 5 P (X = 5) = p (1 − p)4 5 Question b) Y the number of tosses for the first head is geometrically distributed, thus P (Y = 7) = (1 − p)6 p

Question c) Z the number of tosses to get 5 heads follows a negative binomial distribution   11 P (Z = 12) = (1 − p)7 p5 4 Question d) X1 the number of heads in the first 8 tosses and X2 the number of

IMM - DTU

02405 Probability 2004-3-3 BFN/bfn

Question a) X the number of heads in 9 tosses is binominally distributed, thus   9 5 P (X = 5) = p (1 − p)4 5 Question b) Y the number of tosses for the first head is geometrically distributed, thus P (Y = 7) = (1 − p)6 p

Question c) Z the number of tosses to get 5 heads follows a negative binomial distribution   11 P (Z = 12) = (1 − p)7 p5 4 Question d) X1 the number of heads in the first 8 tosses and X2 the number of heads in the next 5 tosses are

IMM - DTU

02405 Probability 2004-3-3 BFN/bfn

Question a) X the number of heads in 9 tosses is binominally distributed, thus   9 5 P (X = 5) = p (1 − p)4 5 Question b) Y the number of tosses for the first head is geometrically distributed, thus P (Y = 7) = (1 − p)6 p

Question c) Z the number of tosses to get 5 heads follows a negative binomial distribution   11 P (Z = 12) = (1 − p)7 p5 4 Question d) X1 the number of heads in the first 8 tosses and X2 the number of heads in the next 5 tosses are independent. We get

02405 Probability 2004-3-3 BFN/bfn

IMM - DTU

Question a) X the number of heads in 9 tosses is binominally distributed, thus   9 5 P (X = 5) = p (1 − p)4 5 Question b) Y the number of tosses for the first head is geometrically distributed, thus P (Y = 7) = (1 − p)6 p

Question c) Z the number of tosses to get 5 heads follows a negative binomial distribution   11 P (Z = 12) = (1 − p)7 p5 4 Question d) X1 the number of heads in the first 8 tosses and X2 the number of heads in the next 5 tosses are independent. We get 5   X 8 i=0

i

i

p (1 − p)

8−i

  5 i p (1 − p)5−i = i

02405 Probability 2004-3-3 BFN/bfn

IMM - DTU

Question a) X the number of heads in 9 tosses is binominally distributed, thus   9 5 P (X = 5) = p (1 − p)4 5 Question b) Y the number of tosses for the first head is geometrically distributed, thus P (Y = 7) = (1 − p)6 p

Question c) Z the number of tosses to get 5 heads follows a negative binomial distribution   11 P (Z = 12) = (1 − p)7 p5 4 Question d) X1 the number of heads in the first 8 tosses and X2 the number of heads in the next 5 tosses are independent. We get 5   X 8 i=0

i

i

p (1 − p)

8−i

  5    X 5 i 8 5 2i 5−i p (1 − p) = p (1 − p)13−2i i i i i=0

02405 Probability 2004-3-3 BFN/bfn

IMM - DTU

Question a) X the number of heads in 9 tosses is binominally distributed, thus   9 5 P (X = 5) = p (1 − p)4 5 Question b) Y the number of tosses for the first head is geometrically distributed, thus P (Y = 7) = (1 − p)6 p

Question c) Z the number of tosses to get 5 heads follows a negative binomial distribution   11 P (Z = 12) = (1 − p)7 p5 4 Question d) X1 the number of heads in the first 8 tosses and X2 the number of heads in the next 5 tosses are independent. We get 5   X 8 i=0

i

i

p (1 − p)

8−i

  5    X 5 i 8 5 2i 5−i p (1 − p) = p (1 − p)13−2i i i i i=0