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S2 - Binomial and Poisson Distributions
Having trouble with a couple of questions in the review exercise. Could some people help me on these questions. Thanks.
22) All the letters in a particular office are typed either by Pat, a trainee typist, or by Lyn, who is a fully-trained typist. The probability that a letter typed by Pat will contain one or more errors is 0.3. Find the probability that a random sample of 4 letters typed by Pat will include exactly one letter free from error.
The probability that a letter typed by Lyn will contain one or more errors is 0.05. Use tables, or otherwise, to find, to 3 decimal places, the probability that in a random sample of 20 letters typed by Lyn, not more than 2 letters will contain one or more errors.
On any one day, 6% of the letters in the office are typed by Pat. One letter is chosen at random from those typed on that day. Show that the probability that it will contain one or more errors is 0.065.
Given that each of 2 letters chosen at random from the day's typing contains one or more errors, find, to 4 decimal places, the probability that one was typed by Pat and the other by Lyn.
25) A process for making plate glass produces small bubbles (imperfections) scattered at random in the glass, at an average rate of four small bubles per 10m^2.
Assuming a Poisson model for the number of small bubbles, determine, to 3 decimal places, the probability that a piece of glass 2.2m x 3.0m will contain:
a) exactly two small bubbles
b) at least one small bubble
c) at most two small bubbles
Find, to 3 decimal places, the probability that five pieces of glass, each 2.5m x 2.0m, will contain a total of at least ten small bubbles.
Answers
22) 0.0756; 0.925; 0.4005
25) (a) 0.249 (b) 0.929 (c) 0.508; 0.542
22) All the letters in a particular office are typed either by Pat, a trainee typist, or by Lyn, who is a fully-trained typist. The probability that a letter typed by Pat will contain one or more errors is 0.3. Find the probability that a random sample of 4 letters typed by Pat will include exactly one letter free from error.
The probability that a letter typed by Lyn will contain one or more errors is 0.05. Use tables, or otherwise, to find, to 3 decimal places, the probability that in a random sample of 20 letters typed by Lyn, not more than 2 letters will contain one or more errors.
On any one day, 6% of the letters in the office are typed by Pat. One letter is chosen at random from those typed on that day. Show that the probability that it will contain one or more errors is 0.065.
Given that each of 2 letters chosen at random from the day's typing contains one or more errors, find, to 4 decimal places, the probability that one was typed by Pat and the other by Lyn.
25) A process for making plate glass produces small bubbles (imperfections) scattered at random in the glass, at an average rate of four small bubles per 10m^2.
Assuming a Poisson model for the number of small bubbles, determine, to 3 decimal places, the probability that a piece of glass 2.2m x 3.0m will contain:
a) exactly two small bubbles
b) at least one small bubble
c) at most two small bubbles
Find, to 3 decimal places, the probability that five pieces of glass, each 2.5m x 2.0m, will contain a total of at least ten small bubbles.
Answers
22) 0.0756; 0.925; 0.4005
25) (a) 0.249 (b) 0.929 (c) 0.508; 0.542
(22)
X
= Number of error-free letters from Pat
~ Bin(4, 0.7)
P(X = 1)
= 4C1 (0.7) (0.3)^3
= 0.0756
--
Y
= Number of error-free letters from Lyn
~ Bin(20, 0.95)
P(Y >= 18)
= 20C18 (0.95)^18 (0.05)^2 + 20C19 (0.95)^19 (0.05) + (0.95)^20
= 0.925
--
P(contains error)
= P(contains error | typed by Pat)*P(typed by Pat) + P(contains error | typed by Lyn)*P(typed by Lyn)
= 0.3*0.06 + 0.05*0.94 . . . . . you can also get this from a tree diagram
= 0.065
--
Let A = {both letters contain an error} and B = {one letter was typed by Pat and one by Lyn}. Then
P(A) = 0.065^2 = 0.004225
P(A and B)
= 2*0.06*0.94*0.3*0.05
= 0.001692
P(B | A)
= P(A and B) / P(B)
= 0.001692 / 0.004225
= 0.400
--
(25)
X
= Number of bubbles in a 2.2m x 3.0m piece of glass
~ Po(4*(2.2*3/10))
= Po(2.64)
P(X = 2)
= e^(-2.64) (2.64)^2/2
= 0.249
P(X >= 1)
= 1 - P(X = 0)
= 1 - e^(-2.64)
= 0.929
P(X <= 2)
= e^(-2.64) (1 + 2.64 + (2.64)^2/2)
= 0.508
--
Y
= Total number of bubbles in five 2.5m x 2.0m pieces of glass
~ Po(5*4*(2.5*2/10))
= Po(10)
P(Y >= 10)
= 1 - P(Y <= 9)
= 1 - e^(-10) (1 + 10 + 10^2/2! + 10^3/3! + ... + 10^9/9!)
= 0.542
X
= Number of error-free letters from Pat
~ Bin(4, 0.7)
P(X = 1)
= 4C1 (0.7) (0.3)^3
= 0.0756
--
Y
= Number of error-free letters from Lyn
~ Bin(20, 0.95)
P(Y >= 18)
= 20C18 (0.95)^18 (0.05)^2 + 20C19 (0.95)^19 (0.05) + (0.95)^20
= 0.925
--
P(contains error)
= P(contains error | typed by Pat)*P(typed by Pat) + P(contains error | typed by Lyn)*P(typed by Lyn)
= 0.3*0.06 + 0.05*0.94 . . . . . you can also get this from a tree diagram
= 0.065
--
Let A = {both letters contain an error} and B = {one letter was typed by Pat and one by Lyn}. Then
P(A) = 0.065^2 = 0.004225
P(A and B)
= 2*0.06*0.94*0.3*0.05
= 0.001692
P(B | A)
= P(A and B) / P(B)
= 0.001692 / 0.004225
= 0.400
--
(25)
X
= Number of bubbles in a 2.2m x 3.0m piece of glass
~ Po(4*(2.2*3/10))
= Po(2.64)
P(X = 2)
= e^(-2.64) (2.64)^2/2
= 0.249
P(X >= 1)
= 1 - P(X = 0)
= 1 - e^(-2.64)
= 0.929
P(X <= 2)
= e^(-2.64) (1 + 2.64 + (2.64)^2/2)
= 0.508
--
Y
= Total number of bubbles in five 2.5m x 2.0m pieces of glass
~ Po(5*4*(2.5*2/10))
= Po(10)
P(Y >= 10)
= 1 - P(Y <= 9)
= 1 - e^(-10) (1 + 10 + 10^2/2! + 10^3/3! + ... + 10^9/9!)
= 0.542
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