third and fourth exercises Homework in probability theory Variant 1 A box contai

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third and fourth exercises

Homework in probability theory Variant 1 A box contains 15 red balls, 10 blue balls, and 5 white balls. l. 2 balls are taken from the box. Consider the following events: D) the second ball is red A) the first ball is red B) the first ball is blue E) the second ball is blue C) the first ball is white F) the second ball is white Find the following probabilities a) p (A) b) p(B) c) p(C) d) p(D/A) e) p (D/B) f p (D/C) k) p(F/B) 1) p(F/C) h) p(E/B) i) p (E/C) j) p(F/A) g) p(E/A) 2. Find probabilities to take c) one blue ball and one white ball a) 2 red balls b) 2 blue balls 3. We have taken 2 balls form the box. What are probabilities that we have taken a) at least one red ball b) at least one blue ball c) at least one white ball 4. We have components manufactured by 3 different factories. 60% are manufactured by 1st factory, 30% by 2nd factory, and 10% by 3rd factory. 1st factory gives 0.5% of nonstandard components, 2nd factory 1%, and 3rd factory 5%. We take one of these components. a) Find probability that component we have taken does not correspond to standards. Component really does not correspond to standards. Find probability that it is manufactured c) by 2nd factory d) by 3rd factory b) by 1st factory

Explanation / Answer

Q.3 There are 15 Red balls , 10 blue balls and 5 white balls. ( 15R, 10B ,5W) out of which two balls are selected

(a) Probability of At least one red ball = Pr ( one red) + Pr( 2 Red)

Pr (One red) = 15 C1 * 15 C1 / 30C2 = 15 * 15 *2 / (30 * 29) = 0.517

Pr (2 Red) = 15 C2 * 15 C0 / 30C2 = 15 * 14/ (30 * 29) = 0.2414

Pr ( at least one red ball) = 0.517 + 0.241 = 0.758

(b) Pr (at least one blue ball) = Pr ( one blue) + Pr( 2 blue)

Pr (One blue) = 10 C1 * 20 C1 / 30C2 = 10 * 20 *2 / (30 * 29) = 0.460

Pr (two blue) = 10 C2 * 20 C0 / 30C2 = 10 * 9/ (30 * 29) = 0.103

Pr ( at least one red ball) = 0.460 + 0.103 = 0.563

(c) Pr (at least one white ball) = Pr ( one white) + Pr( two white)

Pr (One white) = 5 C1 * 25 C1 / 30C2 = 5* 25 *2 / (30 * 29) = 0.287

Pr (two blue) = 5 C2 * 25 C0 / 30C2 = 5 *4/ (30 * 29) = 0.024

Pr ( at least one red ball) = 0.287+ 0.024 = 0.311

Q.4 Pr ( Manurfactured from factory 1 ) = P (F1) = 0.6

similarly, P(F2)= 0.3 and P(F3) = 0.1

Percentage of Non standard componenets by factory 1 = P (NSF1) = 0.005

Percentage of Non standard componenets by factory 2 = P (NSF2) = 0.01

Percentage of Non standard componenets by factory 3 = P (NSF3) = 0.05

(a) Probability that a given componenet is not correspond to standards P(F) = 0.6 * 0.005 + 0.3 * 0.01 + 0.1 * 0.05 = 0.011

so 1.1% percentage chance that a given random component is faulty.

(b) A component doesn't conform to standard so probability that it is manufactured by

(i) by 1st factory = P(F1)* P(NSF1) / P(F) = 0.6 * 0.005/ 0.011 = 0.273 so 27.3 % probability that it belongs to factory 1.

(ii) by 2nd factory =  P(F2)* P(NSF2) / P(F) = 0.3 * 0.01 / 0.011 = 0.273 so 27.3 % probability

(iii) by 3rd factory = P(F3)* P(NSF3) / P(F) = 0.1 * 0.05 / 0.011 = 0.455 so 45.5 % probability

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