Use Bernoulli Binomial Geometric Poisson or other type of sp

Use Bernoulli, Binomial, Geometric, Poisson, or other type of special distribution

(1) Ten percent of computer parts produced by a certain supplier are defective. What is the probability that a sample of 10 parts contains more than 3 defective ones? (2) On the average, two tornadoes hit major U.S. metropolitan areas every year. What is the probability that more than five tornadoes occur in major U.S. metropolitan areas next year? (3) A lab network consisting of 20 computers was attacked by a computer virus. This virus enters each computer with probability 0.4, independently of other computers. a) Find the probability that the virus enters at least 10 computers. b) A computer manager checks the lab computers, one after another, to see if they were infected by the virus. What is the probability that she has to test at least 6 computers to find the first infected one? (4) On the average, 1 computer in 800 crashes during a severe thunderstorm. A certain company had 4,000 working computers when the area was hit by a severe thunderstorm. a) Compute the expected value and variance of the number of crashed computers. b) Compute the probability that less than 10 computers crashed. c) Compute the probability that exactly 10 computers crashed. (You may use a suitable approximation.) (5) A baker put 500 raisins into dough. mixed well, and made 100 cookies. You take a random cookie. What is the probability of finding at least 4 raisins in it? (Hint: use Poisson approximation to the Binomial distribution of the number of raisins in one cookie.)

Solution

(1) Given X follows Binomial distribution with n=10 and p=0.1

P(X=x)=10Cx*(0.1^x)*(0.9^(10-x)) for x=0,1,2,...,10

So the probability is

P(X>3) = 1-P(X=0)-P(X=1)-P(X=2)-P(X=3)

=1-10C0*(0.1^0)*(0.9^(10-0))-...-10C3*(0.1^3)*(0.9^(10-3))

= 0.0127952

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(2)Given X follows Poisson distribution with mean=2

P(X=x)=(2^x)*exp(-2)/x!

So the probability is

P(X>5) =1-P(X=0)-P(X=1)-...-P(X=5)

=1-(2^0)*exp(-2)/1-...-(2^5)*exp(-2)/5!

=0.01656361

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(3)(a) Given X follows Binomial distribution with n=20 and p=0.4

P(X=x)=20Cx*(0.4^x)*(0.6^(20-x)) for x=0,1,2,...,20

So the probability is

P(X>=10) = P(X=10)+P(X=11)+...+P(X=20) =0.2446628

(b) X follows Geometric distribution with p=0.4

P(at least 6) = 1-P(X=1)-P(X=2)-...-P(X=5)

=1-0.4 -0.6*0.4-0.6^2*0.4-0.6^3*0.4-0.6^4*0.4

=0.07776

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(4)(a) expected value= n*p=4000*(1/800) = 5

variance= n*p*(1-p) =4000*(1/800)*(1-1/800) =4.99375

(b)P(X<10) = P((X-mean)/s <(10-5)/sqrt(4.99375))

=P(Z<2.24) =0.9875 (from standard normal table)

(c) P(X=10) = P(9.5<X<10.5)

=P((9.5-5)/sqrt(4.99375)<Z<(10.5-5)/sqrt(4.99375))

=P(2.01<Z<2.46) =0.0153 (from standard normal table)

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(5)Given X follows Poisson distribution with mean=n*p=100*(1/500)=0.2

P(X=x)=0.2^x*exp(-0.2)/x!

So the probability is

P(X>=4)=1-P(X=0)-P(X=1)-...-P(X=3)= 0.0000568