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Consider the binomial distribution with n trials and P(S) = p (a) Show that p(v). n -y + 1)P for y= 1.2. Ply- 1) .., n. Equivalently, for y = 1, 2, .., n, the equation ply) = - y - 1) P ply - 1) gives a recursive relationship between the probabilities associated with successive values of Y. ya Py – 1) -- 1) - n-y+ 1)p ya (b) If n= 70 and p= 0.04, use the above relationship to find P(Y < 3). (Round your answer to four decimal places.) (e) Show that ply) Ply – P)- -y + 1) P > 1 if y < (n + 1)p. This establishes that ply) > Ply - 1) if y is small (y < (n - 1)p) and ply) < ply - 1) if y is large (y > (n - 1) p). Thus, successive binomial probabilisies increase for a while and decrease from then on. (n + 1)p > y (n -1- y)p > y- (n -1- y)p > (n-1-ye > 1 ya Ply - 1) 1ify > (n - 1) p. (n + 1)p < y Show thatP) (n +1- y)p < y- (n -1- y)p < (n+1-y < 1 yq Show that PV) =1 if (n + 1)lp is an integer and y = (n + 1)p. Ply – 1) (n + 1)p = y (n +1- y)p - y+ (n +1- y)p = (n +1- yle- 1

Consider the binomial distribution with n trials and P(S) = p (a) Show that p(v). n -y + 1)P for y= 1.2. Ply- 1) .., n. Equivalently, for y = 1, 2, .., n, the equation ply) = - y - 1) P ply - 1) gives a recursive relationship between the probabilities associated with successive values of Y. ya Py – 1) -- 1) - n-y+ 1)p ya (b) If n= 70 and p= 0.04, use the above relationship to find P(Y < 3). (Round your answer to four decimal places.) (e) Show that ply) Ply – P)- -y + 1) P > 1 if y < (n + 1)p. This establishes that ply) > Ply - 1) if y is small (y < (n - 1)p) and ply) < ply - 1) if y is large (y > (n - 1) p). Thus, successive binomial probabilisies increase for a while and decrease from then on. (n + 1)p > y (n -1- y)p > y- (n -1- y)p > (n-1-ye > 1 ya Ply - 1) 1ify > (n - 1) p. (n + 1)p < y Show thatP) (n +1- y)p < y- (n -1- y)p < (n+1-y < 1 yq Show that PV) =1 if (n + 1)lp is an integer and y = (n + 1)p. Ply – 1) (n + 1)p = y (n +1- y)p - y+ (n +1- y)p = (n +1- yle- 1

A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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Consider the binomial distribution with n trials and P(S) = p (a) Show that _PLY) = n -y+ 1)P for y = 1, 2. Ply - 1) ... n. Equivalently, for y = 1, 2, .., n, the equation p(y) = -y + 1)P p(y - 1) gives a recursive relationship between the probabilities associated with successive values of Y. ya yg Ply - 1) (,": Jo - - ( - 1) - n- y+ 1)p ya (b) Ifn = 70 and p = 0.04, use the above relationship to find P(Y < 3). (Round your answer to four decimal places.) (c) Show that – P = -Y* L)P > 1 if y < (n + 1)p. This establishes that p(y) > P(v - 1) if y is small (y < (n + 1)p) and ply) < p(y - 1) if y is large (y > (n + 1) p). Thus, successive binomial probabilities increase for a while and decrease from then on. Ply - 1) (n + 1)p > y yq (n +1- y)p > y+ (n +1- y)p > (n +1- y)p > 1 yq Show that PV) <1 if y > (n + 1) p. Ply - (n + 1)p < y (n +1- y)p « y - (n +1- y)p < (n +1- y)e < 1 yq Show that PL = 1 if (n + 1)p is an integer and y = (n + 1)p. Ply - 1) (n + 1)p = y (n +1- y)p = y + (n +1- y)p = (n +1- yle = 1 ya
Transcribed Image Text:Consider the binomial distribution with n trials and P(S) = p (a) Show that _PLY) = n -y+ 1)P for y = 1, 2. Ply - 1) ... n. Equivalently, for y = 1, 2, .., n, the equation p(y) = -y + 1)P p(y - 1) gives a recursive relationship between the probabilities associated with successive values of Y. ya yg Ply - 1) (,": Jo - - ( - 1) - n- y+ 1)p ya (b) Ifn = 70 and p = 0.04, use the above relationship to find P(Y < 3). (Round your answer to four decimal places.) (c) Show that – P = -Y* L)P > 1 if y < (n + 1)p. This establishes that p(y) > P(v - 1) if y is small (y < (n + 1)p) and ply) < p(y - 1) if y is large (y > (n + 1) p). Thus, successive binomial probabilities increase for a while and decrease from then on. Ply - 1) (n + 1)p > y yq (n +1- y)p > y+ (n +1- y)p > (n +1- y)p > 1 yq Show that PV) <1 if y > (n + 1) p. Ply - (n + 1)p < y (n +1- y)p « y - (n +1- y)p < (n +1- y)e < 1 yq Show that PL = 1 if (n + 1)p is an integer and y = (n + 1)p. Ply - 1) (n + 1)p = y (n +1- y)p = y + (n +1- y)p = (n +1- yle = 1 ya
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