Enterprise Solution Division Random Process In a random process we know that what outcomes or events could happen; but we do not know which particular outcome or event will happen. For example tossing of coin‚ rolling of dice‚ roulette wheel‚ changes in valuation in shares‚ demand of particular product etc. Probability It is the numeric value representing the chance‚ likelihood‚ or possibility a particular event will occur It is measured as the fraction between 0 & 1 (or 0% &100%) Probability
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a particular event will happen if something is done repeatedly‚ (596 Webster’s Dictionary). You cannot determine any events that will happen in the future‚ because there is always a chance that something odd will happen‚ (Linn 39-40). Probability originally started for the purpose and attempt to analyze games of chance. Probability is also used in determining the outcomes of an experiment. Sample space is the collection of all results. Probability is a way to assign every event a value between
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the likelihood of an event happening. Directly or indirectly‚ probability plays a role in all activities. Probability is a measure or estimation of how likely it is that something will happen or that a statement is true. Probabilities are given a value between 0 (0% chance or will not happen) and 1 (100% chance or will happen). The higher the degree of probability‚ the more likely the event is to happen‚ or‚ in a longer series of samples‚ the greater the number of times such event is expected to happen
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coin is tossed (H‚ T). (a) Enumerate the elementary events in the sample space for the die/coin combination. (b) Are the elementary events equally likely? Explain. A) Elementary events are - DIE COIN 1 2 3 4 5 6 HEADS H1 H2 H3 H4 H5 H6 TAILS T1 T2 T3 T4 T5 T6 B) YES‚ EACH EVENT IS EQUALLY LIKELY TO OCCUR. THERE ARE 12 POSSIBLE OUTCOMES AS A RESULT OF ROLLING OE DIE AND FLIPPING ONE COIN‚ THEREFORE THE LIKELYHOOD OF ANY ONE EVENT OCCURING IS 1/12. 5.13 (page 186)‚ 5.13 Given P(A)
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of the basic rules of probability: the sum of the probability that an event will happen and the probability that the event won’t happen is always 1. (In other words‚ the chance that anything might or might not happen is always 100%). If we can work out the probability that no two people will have the same birthday‚ we can use this rule to find the probability that two people will share a birthday: P(event happens) + P(event doesn’t happen) = 1 P(two people share birthday) + P(no two people share
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13 4. Introduction to Probability ....................................................................... 15 5. Unions‚ Intersections‚ and Complements ................................................ 23 6. Conditional Probability & Independent Events..................................... 28 7. Discrete Random Variables....................................................................... 33 8. Binomial Random Variable ...................................................................... 37
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Many animals‚ despite being from different species‚ create mutual bonds like no other. The same could be said about a dog that just gave birth‚ may take in a litter of kittens. In the passage‚ “Animal Roles and Relationships‚” we read about different animal relationships. While they are from different species‚ they can come together for the benefit of both species‚ and themselves as well. An example of this is the hermit crab and the sea anemone. “Hermit crabs use their pincers to tap sea anemones
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weights. Then 6 1 ‚...‚m(ω6 ) = 21 . (Check for yourself that this choice of values of m(ωi ) satisfies m(ω1 ) = 21 the three conditions above!) Therefore‚ P (Even) = P ({2‚ 4‚ 6} = 2 21 + 4 21 + 6 21 = 12 21 = 4 7 = 0.57. 7. Let A and B be events such that P (A ∩ B) = 14 ‚ P (Ac ) = 13 ‚ and P (B = 12 . What is P (A ∪ B)? Recall Theorem 4 from class: P (A ∪ B) = P (A) + P (B) − P (A ∩ B). We already know that P (B) = 12 and P (A ∩ B) = 14 ‚ so we just need to find P (A). By Theorem 1 part
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shift requires 6 operators‚ 2 maintenance persons‚ and 1 supervisor‚ in how many different ways can it be staffed? [8 points] 18 10 4 18564 45 4 3‚341‚520 6 2 1 (b) Suppose A and B are not mutually exclusive events‚ and we have P(A)=0.35‚ P(B)=0.40‚ P(AB)=0.18. Compute the following probabilities: i) P (AB)=? [4 points] P (AB)=P[A]+P[B]-P[AB]=0.35+0.40-0.18=0.57 ii) P(AB)=? P[A B] P[ A B] 0.18 0.45 P[ B] 0.40 1 of 6 [4 points] Name: Problem
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input the probabilities of 0.4‚ 0.4 and 0.2 for “good”‚ “moderate” and “poor” market reception. We then proceed to develop the marginal‚ conditional‚ and joint probabilities for each terminal end-point. The formula for the conditional probability of events A and B is changed as: P(A ∩ B) = P(B) P(A | B) By developing the likely revenue of market response outcome and summing the results‚ we obtain the expected
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