# (a) If chances of success in a contract are 15% for a contractor in Government tenders and if he has quoted in 5 tenders recently, find the probabilities that, (i) He does not get any contract...

(a) If chances of success in a contract are 15% for a contractor in Government tenders and if he has quoted in 5 tenders recently, find the probabilities that,

(i) He does not get any contract

(ii) He gets only one contract

(iii) He gets at least one contract, using Binomial Distribution

(b) A Japanese car manufacturer claims that on an average there are just 21 defects per 1000 cars manufactured by it. If the no of defects follow Poisson’s distribution, find the probabilities:

A car selected randomly from the recently produced lot has no defect

A car selected randomly has exactly two defects

A car selected randomly has at least two defects

*print*Print*list*Cite

Start by breaking the the given of the first problem. It says that success to get the contract is 15 %.

Let p the success

q the failure

n the number of tries.

So p =0.15

q = 0.85

n = 5

To solve for the probability of getting no contract (i)

(i) 1 - probability of success of all tries.

(i) `P(0) = 1 - (0.15)^5`

(i) `P(0) = 0.9999`

To solve for probability that he only gets 1 contract,

(ii) `P(1) = nC_1 * p^1 * q^(n-1)`

(ii) `P(1) = 5C_1 * 0.15^1 * 0.85^4`

(ii) `P(1) = 0.3915`

To solve for the probability that he gets atleast 1,

(iii) `P(le1) = P(1) + P(2) + P(3) + P(4) + P(5)`

(iii) `P(le 1) =5C_1 * 0.15^1 * 0.85^4 +5C_2 * 0.15^2 * 0.85^3 +5C_3 * 0.15^3 * 0.85^2 +5C_4 * 0.15^4 * 0.85^1 +5C_5 * 0.15^5 * 0.85^0`

(iii) `P(le 1) = 0.5563`

There is a correction for (iii), the notation should `P(>= 1).`

That's the only error, the answer is correct though.

For the secon problem, use Poisson Distribution Formula:

`f(k,lambda) = ((lambda)^k *e^(-lambda))/(k!)`

where k is the occurences of events,

`lambda` is a positive real number

`k =0, 1, ......`

`lambda = 21`

no defects, k=0;

`P (k = 0) = (21^0*e^-21)/(0!)`

has 2 defects, k = 2

`P(k=2) = (21^2*e^-2)/(2!)`

`P(k = 2) = 1.6720 x 10^-7`

has atleast 2 defects, k >= 2

`P(k>=2) =sum_(i=1)^k ((lambda)^i*e^(-lambda))/(i!)`

k =21

`P(k>=2) = sum_(i=1)^k (21^i*e^-21)/(i!)`

`P(k>=2) = 0.5577`

From a large consignment of apples, 400 apples were selected randomly and out of these 65 apples were found to be defective. Find the Standard Error of proportion of bad apples in the consignment. Also determine the range of proportion within which the population proportion would lie with 95% confidence .