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09 Midterm Review-1

# In a given week the probability of facing a

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In a given week, the probability of facing a disruption is 5% for the Malaysian supplier and 7% for the Mexican supplier. The suppliers can independently face disruptions.

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Example 22 a) What is the probability that both suppliers will face a disruption, and the manufacturer will not be able to source its critical component during a given week? b) What is the probability that only one supplier faces a disruption, and the manufacturer has one supplier to depend on for critical component? c) The manufacturer is told that one of the suppliers has faced a disruption. What is the probability that the disrupted supplier is located in Malaysia? d) The manufacturer has an option to source from a third supplier, as well, whose probability of facing a disruption during a week is 8%. What is the probability that the manufacturer can now guarantee sourcing the critical component in a
Example 23 a) Because facing a disruption is independent across suppliers, P(MaMe)=P(Ma) P(Me)= .05*.07=.0035=0.35%. Thus, the probability that both suppliers cannot supply components in a given week is very low. b) Probability that only Malaysian supplier is disrupted = P(MaMe’)=P(Ma) P(Me’)= 0.05 * (1-.07) = 0.0465. Probability that only Mexican supplier is disrupted = P(Ma’Me)=P(Ma’) P(Me’)= (1-.05) *.07 = 0.0665. We add up the two probabilities and obtain 0.0465+0.0665 = 0.113 =11.3% as the probability that one supplier is disrupted, and one supplier is not. The manufacturer is told that one of the suppliers has faced a disruption. c) We need to compute P(Ma | only one supplier is disrupted) P(Ma  one supplier is disrupted) / P(only one supplier is disrupted)= P(Ma  Me’) / P(only one supplier is disrupted)= 0.0465 / 0.113=41.15% d) Probability that all suppliers are disrupted at the same time is . 05*.07*.08=0.00028. With the remaining probability, 1- 0.00028=0.99972=99.972%, there is at least one supplier, who is not disrupted, hence, the manufacturer has guaranteed supply of components.

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Discrete Probability Distributions 24 Discrete random variable: when there are finitely many or countable number of values of a random variable Continuous random variable: when a random variable can take any real-number value (in an open or closed range) Probability mass function for discrete random variables p(x) = Probability the random variable X takes on the value x p(x) ≥ 0 for all x and p(x) = 1 Cumulative distribution function (CDF): the probability that a random variable X takes on a value less than or equal to x. F ( x ) = Pr( X x ) = p ( i ) i x
25 Discrete Probability Distributions Mean of a probability distribution: A probability-weighted average of possible outcomes. Variance of a random variable X is the weighted sum of its squared deviation from its mean Standard deviation: the square root of the variance where E[X]: Expected value of x x: Values of the random variable P(x): Probability of the random variable taking on the value x.

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In a given week the probability of facing a disruption is 5...

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