# (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

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...

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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`

Approved by eNotes Editorial Team

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`

Approved by eNotes Editorial Team