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