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Probability

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Discussion Topics Conditional Probability Probability using the general multiplication rule Independent event

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Question 1 Males (In millions) Females (In millions) Totals (In millions) Never married30.325.055.3 Married63.664.1127.7 Widowed2.611.313.9 Divorce9.713.122.8 Totals106.2113.5219.7 Calculate the probability that a randomly selected individual 18 years or older is widowed Calculate the probability that a randomly selected individual 18 years or older is widowed and female The data in the table below represent the marital status of males and females 18 years old or older

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Solution Probability that selected individual is widowed: (a)Define W as the event of picking a widow P(W)=Total of widows/Overall totals F6/F8

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Solution Probability that selected individual is widowed and female P(Widowed | Female)=Number of widowed/Number of females E6/E8

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Question 2 In the game of roulette, the wheel has slots numbered 0,00, and 1 through 36.A metal ball is allowed to roll around a wheel until it falls into one of the numbered slots. You decide to play the game and place a bet on the number 17. (a) What is the probability that the ball will land in the slot numbered 17 two times in a row?

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Solution The events are independent because the probability of the 1 st spin cannot affect the probability of the second spin. The ball does not remember it landed on 17 on the first spin (1/38)/(1/38)

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Question 3 Suppose that a box of 100 circuits is sent to a manufacturing plant. Of the 100 circuits shipped, 5 are defective. The plant manager receiving the circuits randomly selects 2 and tests them. If both circuits work, she will accept the shipment. Otherwise the shipment is rejected. (a)Use a tree diagram to represent all possible outcomes (b) What is the probability that the plant manager discovers at least 1 defective circuit and rejects the shipment

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Tree diagram Tool used is Microsoft Visio.

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From our tree diagram and using addition rule: P(at least 1 defective) = P(GD) + P(DG) + P(DD) = 0.048+0.048+0.002 = 0.098 There is 0.098 probability of shipment being rejected 2 nd Approach: P(at least 1 defective) = 1-P(none defective) = 1- (95/100)*(94/99) = 1- 0.902 = 0.098

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CSTEM Web link http://www.cis.famu.edu/~cdellor/math/ Presentation slides Problems Guidelines Learning materials

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Intro to Probability Section 4.2. PROBABILITY All probabilities must be between 0 and 1.

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