Question

Sam is writing a research paper for his graduation. On his draft of the paper, there are an average of 5 typos on each page. Assume each typo is independent.

d) What is the probability that Sam has 10 typos in Chapter 1 and 10 typos in Chapter 8 (which is 5 pages)?

e) It turns out that the first 4 chapters have the same number of pages (4 pages), what is the probability that Sam has 18 to 20 typos (inclusive) in 2 of the 4 chapters of his draft?

Answer 1

The number of typos per page = 5.

d) Chapter 1 contains 5 pages.

X : number of typos in chapter 1

Since number of typos per page is 5 and each typo is independent.

Hence the random variable X follow Poisson distribution with parameter lambda = 5 * 5 = 25.

The probability distribution of X is

Chapter 8 contains 5 pages.

Y : number of typos in chapter 8 .

Since number of typos per page is 5 and each typo is independent.

Hence the random variable Y follow Poisson distribution with parameter lambda = 5 * 5 = 25.

Required Probability = P ( X =10 , Y =10)

Since X and Y are independent

P ( X =10 , Y =10) = P ( X = 10) * P(Y=10)

Required Probability = 0.0004* 0.0004

= 1.6 * 10 ^-7.

P ( Sam has 10 typos in chapter 1 and 10 typos in chapter 8 ) = 1.6 * 10 ^-7.

e) Since each chapter contain 4 pages and number of typos per page is 5.

Consider the random variable

W: Number of typos per chapter

The probability distribution of random variable W is Poisson with parameter lambda = 4*5 = 20

The probability distribution of W is

P ( Chapter contains 18-20 typos) = P(W=18) + P(W=19) + P(W=20)

P ( Chapter contains 18-20 typos) = 0.0844 + 0.0888 + 0.0888 = 0.2620

n = number of chapter = 4.

p = probability of each chapter contains 18 to 20 typos. = 0.2620

Consider Z : Number of chapter contains 18 to 20 typos.

The probability distribution of Z is binomial with n = 4 and p = 0.2620.

The probability distribution of Z is

Required Probability = P ( Z=2)

=0.2243

P ( Sam has 18 to 20 typos in the 2 of the 4 chapter of his draft) = 0.2243.