Probability: Playing Card and Vocational Training Essay

Q1). You draw a card at random from a standard deck of 52 cards. Neither you nor anyone else looked at the card you picked. You keep it face down. Your friend then picks a card at random from a remaining 51 cards. a) What is the probability that your card is ace of spades? 1/52 b) What is the probability that your friend’s card is ace of spades? (Hint: Construct the sample space for what your friend’s card can be.) 1/51 c) You turn over your card and it is 10 of diamonds. Now what is the probability that your friend’s card is the ace of spades? 1/51 Q2). Suppose that 25% of the population in your area is exposed to a television commercial of Ford Automotive, and 34% is exposed to Ford’s radio advertisements. Also it is known that 10% of the population is exposed to both means of advertising. If a person is randomly chosen out of the entire population in this area, what is the probability that he or she was exposed to at least one of the two modes of advertising? ANS: P(A) = 0.25 P(B) = 0.34 P(A^B) = 0.10. Probability that he/she was exposed to at least one mode of advertising = 0.25 + 0.34 + 0.1 = 69% Q3). A Firm has 550 employees, 380 of them have had at least some college education, and 412 of the employees underwent a vocational training program. Furthermore, 357 employees are both college educated and have had vocational training. If an employee is chosen at random, what is the probability that he or she is college educated or has had the vocational training or both? Q4). A bank loan officer knows that 12% of the bank’s mortgage holders lose their jobs and default on the loan in the course of 5 years. She also knows that 20% of the bank’s mortgage holders lose their jobs during this period. Given that one of her mortgage holders just lost his job, what is the probability that he will now default on the loan? Q5). A Recent survey conducted by Towers Perrin and published in the Financial Times showed that among 460 organizations in 13 European Countries, 93% have bonus plans, 55% have cafeteria-style benefits, and 70% employ home-based workers. If the types of benefits are independent, what is the probability that an organization selected at random will have at least one of the three types of benefits? P(A) = 0.93, P(B) = 0.55, P(C) = 0.70. P(Not A) = 1 – 0.93, P(Not B) = 1-0.55, P(Not C)= 1-0.70 P(Not A B C) = 0.07 * 0.45 * 0.30 = 0.01155 P(atleast 1 event occurring) = 1-0.01155 = 0.988.