Mathematics, 21.10.2020 16:01 charethafoster

Five companies (A, B, C, D, and E) that make elec- trical relays compete each year to be the sole sup- plier of relays to a major automobile manufacturer. The auto company’s records show that the probabili- ties of choosing a company to be the sole supplier are Supplier chosen: A B C D E Probability: .20 .25 .15 .30 .10 a. Suppose that supplier E goes out of business this year, leaving the remaining four companies to compete with one another. What are the new probabilities of companies A, B, C, and D being chosen as the sole supplier this year? b. Suppose the auto company narrows the choice of suppliers to companies A and C. What is the probability that company A is chosen this year?

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Answer from: UncleE123

P(a | e') = 0.22

P(b | e') = 0.28

P(c | e') = 0.33

P(a | e' , d' , b') = 0.57

Step-by-step explanation:

From the question we are told that

The probabilities are

Supplier chosen A B C

Probability P(a) = 0.20 P(b) = 0.25 P(c) = 0.15

D E

P(d) = 0.30 P(e) = 0.10

Generally the new probability of companies A being chosen as the sole supplier this year given that supplier E goes out of business is mathematically represented as below according to Bayes theorem

$P(a | e') = \frac{P (a \ and \ e')}{P(e')}$

$P(a | e') = \frac{P (a)}{P(e')}$

$P(a | e') = \frac{P (a)}{1- P(e)}$

=> $P(a | e') = \frac{ 0.20}{1- 0.10}$

=> P(a | e') = 0.22

Generally the new probability of companies B being chosen as the sole supplier this year given that supplier E goes out of business is mathematically represented as below according to Bayes theorem

$P(b | e') = \frac{P (b \ and \ e')}{P(e')}$

$P(b | e') = \frac{P (b)}{P(e')}$

$P(b | e') = \frac{P (b)}{1- P(e)}$

=> $P(b | e') = \frac{ 0.25}{1- 0.10}$

=> P(b | e') = 0.28

Generally the new probability of companies C being chosen as the sole supplier this year given that supplier E goes out of business is mathematically represented as below according to Bayes theorem

$P(c | e') = \frac{P (c \ and \ e')}{P(e')}$

$P(c | e') = \frac{P (c)}{P(e')}$

$P(c | e') = \frac{P (c)}{1- P(e)}$

=> $P(c | e') = \frac{ 0.15}{1- 0.10}$

=> P(c | e') = 0.17

Generally the new probability of companies D being chosen as the sole supplier this year given that supplier E goes out of business is mathematically represented as below according to Bayes theorem

$P(d | e') = \frac{P (d \ and \ e')}{P(e')}$

$P(d | e') = \frac{P (d)}{P(e')}$

$P(d | e') = \frac{P (d)}{1- P(e)}$

=> $P(d | e') = \frac{ 0.30}{1- 0.10}$

=> P(c | e') = 0.33

Generally the probability that B, D , E are not chosen this year is mathematically represented as

P(N) = 1 - [P(e) +P(b) + P(d) ]

=> P(N) = 1 - [0.10 +0.25 +0.30 ]

=> P(N) = 0.35

Generally the probability that A is chosen given that E , D , B are rejected this year is mathematically represented as

$P(a | e' , d' , b') = \frac{P(a)}{P(N)}$

=> $P(a | e' , d' , b') = \frac{0.20 }{0.35 }$

=> P(a | e' , d' , b') = 0.57

Answer from: Quest

here the number -5 is all of the following except a rational number.

step-by-step explanation:

Answer from: Quest

the answer is a

step-by-step explanation:

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