Probability
- 1. 1) You are taking a 10 question multiple choice test. If each question has four choices and you guess on each question, what is the probability of getting exactly 7 questions correct? SOLUTION: n = 10 k = 7 n – k = 3 p = 0.25 = probability of guessing the correct answer on a question q = 0.75 = probability of guessing the wrong answer on a question 2) Births in a hospital occur randomly at an average rate of 1.8 births per hour. What is the probability of observing 4 births in a given hour at the hospital? SOLUTION: (𝑒−1/8∗( 1 8 ) 4 ) 4! 3) Product A is produced by a fabric. In that Fabric a test is made on 100 products. If at least 2 products are defective, the products aren’t put up on sale. What is the probability of putting products up on sale? SOLUTION: ( 100 99 ) ∗ 0,9999 ∗ 0,011 + ( 100 100 ) ∗ 0,99100 ∗ 0,010 4) A representative from the National Football League's Marketing Division randomly selects people on a random street in Kansas City, Kansas until he finds a person who attended the last home football game. Let p, the probability that he succeeds in finding such a person, equal 0.20. And, let X denote the number of people he selects until he finds his first success.
- 2. A) What is the probability that the marketing representative must select 4 people before he finds one who attended the last home football game? SOLUTION: To find the desired probability, we need to find P(X = 4), which can be determined readily using the p.m.f. of a geometric random variable with p = 0.20, 1−p = 0.80, and x = 4: P(X=4)=0,803 ×0.20=0.1024 B) How many people should we expect (that is, what is the average number) the marketing representative needs to select before he finds one who attended the last home football game? And, while we're at it, what is the variance? SOLUTION: The average number is: μ=E(X)= 1 𝑃 = 1 0.20 =5 That is, we should expect the marketing representative to have to select 5 people before he finds one who attended the last football game. Of course, on any given try, it may take 1 person or it may take 10, but 5 is the average number. The variance is 20, as determined by: σ2=Var(X)= 1−𝑃 𝑃2 = 0,80 0,202 =20 5) Suppose that the amount of time one spends in a bank is exponentially distributed with mean 10 minutes, _ = 1/10. What is the probability that a customer will spend more than 15 minutes in the bank? What is the probability that a customer will spend more than 15 minutes in the bank given that he is still in the bank after 10 minutes? SOLUTION: P(X > 15) = 𝑒−15 λ = 𝑒−3/2 = 0.22 P(X > 15|X > 10) = P(X > 5) = 𝑒−1/2 = 0.604 6) The random variable X of the life-lengths of batteries is associated with a probability density function of the form,
- 3. 1 4 𝑒−𝑥/4, x>0 F(x) { 0, elsewhere with measurements in 100 hours. Find the probability the life of a particular battery of this type is greater than 800 hours SOLUTION: P(X> 8) =∫ 1 4 𝑒−𝑥/4 𝑑𝑥 ∞ 8 =−𝑒−𝑥/4 |∞8 =𝑒−2 ≈ 0.135