Question

1. A horse race contains 8 horses. If you wished to bet on every possible top three finishes (first, second, and third place), how many bets would you need to place?
2. The following represents a binomial experiment. Seven plants are operated by a garment manufacturer in seven different countries. There is a 10% chance of strike at any time. It is also believed that strikes at one plant do not influence strikes at any other plant (because these are in third world countries with little access to news information.)  What is the probability of at most two plants going on strike?
3. The pH measurements of water specimens from various locations along a given river basin are normally distributed with mean 8 and standard deviation 0.3.  What is the probability that the pH measurement of a randomly selected water specimen is dangerous—less than 7.2?

Answer 1

The following represents a binomial experiment. Seven plants are operated by a garment manufacturer in seven different countries. There is a 10% chance of strike at any time. It is also believed that strikes at one plant do not influence strikes at any other plant (because these are in third world countries with little access to news information.) What is the probability of at most two plants going on strike?

Answer)

As there are fixed number of trials and probability of each and every trial is same and independent of each other

Here we need to use the binomial formula

P(r) = ncr*(p^r)*(1-p)^n-r

Ncr = n!/(r!*(n-r)!)

N! = N*n-1*n-2*n-3*n-4*n-5........till 1

For example 5! = 5*4*3*2*1

Special case is 0! = 1

P = probability of single trial = 0.1 (10%)

N = number of trials = 7

R = desired success = at most 2 = p(0) + p(1) + p(2)

After substitution requiried probability is p(0) + p(1) + p(2)

= 0.9743085

Question)

The pH measurements of water specimens from various locations along a given river basin are normally distributed with mean 8 and standard deviation 0.3. What is the probability that the pH measurement of a randomly selected water specimen is dangerous—less than 7.2?

Answer)

As the data is normally distributed we can use standard normal z table to estimate the answers

Z = (x-mean)/s.d

Given mean = 8

S.d = 0.3

Z = (7.2 - 8)/0.3 = -2.67

From z table, P(z<-2.67) = 0.0038