The game 'Chunk' is played at carnivals in some parts of Europe.its rules are as follows:- If you pick a number from 1 to 6 and the operator rolls three dice. If the number you picked comes up on all three dice, the operator pays you Rs 3 ; If it comes up on two dice, you are paid Rs 2 ; and it comes up on just one dice, you are paid Rs 1. Only if the number you picked does not come up at all, you pay the operator Rs 1. The probability that you will win money playing in this game is
Answer: C If one picks up a number out of six numbers then the only case in which he/she will loose money if none of the three dice shows the picked number on the top surface. Required probability of loosing the game = 5/6 * 5/6 * 5/6 = 125/216. probability winning the game = 1 - 125/216 = 91/216 = 0.42
Q. No. 14:
A bag contains 10 balls numbered from 0 to 9. the balls are such that the person picking a ball out of the bag is equally likely to pick anyone of them. A person picked a ball and replaced it in the bag after noting its number. He repeated this process 2 more times. What is the probability that the ball picked first is numbered higher than the ball picked second and the ball picked second is numbered higher than the ball picked third?
Answer: B Let the number on the ball picked first = a, second =b and third =c The three numbers a,b and c are distinct. Three distinct ball can be picked in (10*9*8) ways. The order of a,b and c can be as follows:- (i). a>b>c ; (ii) a>c>b ; (iii). b>c>a ; (iv). b>a>c ; (v). c>a>b ; (vi). c>b>a They will occur equal number of times. Thus the number of ways in which (a>b>c)= 1/6*(10*9*8) = 120. Required probability = 120/(10*10*10) = 3/25
Q. No. 15:
A medical clinic test a blood for certain disease from which approximately on person in a hundred suffers. People come to the clinic in group of 50. The operator of the clinic wonders whether he can increase the efficiency of the testing procedure by conducting pooled tests. In the pooled test, the operator would pool the 50 blood samples and test them altogether.If the polled test was negative then he could pronounce the whole group healthy. If not, he could then test each person's blood individually. The expected number of test the operator will have to perform if he pools the blood samples are
Answer: C P(+ test) = 0.01 P(- test) = 0.99 Let X is the random variable representing number of test, and then probability distribution function of X is given by X=1, P(X)=(0.99)50 X= 50, P(X) = (1-(0.99)50)(0.99)49 And for X= 51 , P(X)= (1-(0.99)50)(1-(0.99)49). Then expected number of tests is given by Sigma(xi*P(xi)) => 1*(0.99)50 + 50(1-(0.99)50)*((0.99)49) + 51(1-(0.99)50)(1-(0.99)49) = 20.508. Hence, expected number of tests required is 21.
Q. No. 16:
McDonald’s ran a campaign in which it gave game cards to its customers. These game cards made it possible for customers to win hamburgers, French fries, soft drinks, and other fast-food items, as well as cash prizes. Each card had 10 covered spots that could be uncovered by rubbing them with a coin. Beneath three of these spots were “No Prize” signs. Beneath the other seven spots were names of prizes, two of which were identical. For example, one card might have two pictures of a hamburger, one picture of a Coke, one of French fires, one of a milk shake, one of $5, one of $1000,and three “No Prize” signs. For this card the customer could win a hamburger. To win on any card, the customers had to uncover the two matching spots (which showed the potential prize for that card) before uncovering a “No Prize”; any card with a “No Prize” uncovered was automatically void. Assuming that the two matches and the three “No Prize” signs were arranged randomly on the cards, what is the probability of a customer winning?
Answer: A As per the question there are 10 cover spots out of which (i) Three spots are there with no prize (idemcal) (ii) Two spots of the same sign (Prize) (iii) Five other spots which are distinct Let three spots of no prize are (x, x, x), two spots of same sign are b (P, P) and five other spots are (A, B,C, D, E). Total number of cases without restriction = 10!/(3!*2!) =>Total number of favourable cases happen in the following ways which shows sequence of can covering CASE I:- Second uncovering is P P P _ _ _ _ _ _ _ _ = 8C3*5!*1 CASE II:- _ _ P _ _ _ _ _ _ = 7C3*5!*2 CASE III:-6C3*5!*3 CASE IV:-5C3*5!*4 CASE V:-4C3*5!*5 CASE VI:- 3C3*5!*6 Total number of favourable cases = case (I+II+III+IV+V+VI) / (10!/(3!*2!)) => 0.10
Q. No. 17:
Amit, Sumit and Pramit go to a seaside town to spend a vacation there and on the first day everybodydecides to visit different tourist locations. After breakfast, each of them boards a differenttourist vehicle from the nearest bus-depot.After three hours,Sumit who had gone to a famous beach,calls on the mobile of Pramit and claims that he has observed a shark in the waters. Pramit learnsfrom the local guide that at that time of the year, only eight sea-creatures (including a shark) areobservable and the probability of observing any creature is equal. However, Amit and Pramit laterrecall during their discussion that Sumit has a reputation for not telling the truth five out of six times.What is the probability that Sumit actually observed a shark in the waters?
Answer: A The probability that Sumit actually sees a Shark, given that he claimed to have seen one. => P(He actually sees the shark & reports truth) /P(He claims of seeing a shark) => {P(Sees the shark) × P(Reports Truth)} / {P(sees the shark) × P(reports truth) +P(Doesn't see) × P(reports false)} => {1/8*1/6}/ {(1/8*1.6) + (7/8 *5/6} => 1/36.
Q. No. 18:
Sun Life Insurance company issues standard, preferred and ultra-preferred policies. Among the company's policy holders of a certain age, 50% are standard with the probability of 0.01 dying in the next year, 30% are preferred with a probability of 0.008 of dying in the next year and 20% are ultra-preferred with a probability of 0.007 of dying in the next year. If a policy holder of that age dies in the next year, what is the probability of the decreased being a preferred policy holder ?
Answer: B The percentage of three different types of policy holders and the corresponding probability of dying in the next 1 year are as follow :
Type
Standard
Preferred
Ultra-Preferred
Percentage
50
30
20
Probability
0.01
0.008
0.007
The expected number of deaths among all the policy holders of the given age [say P] during the next year. = T [50*0.01/100 + 30*0.008/100 + 10*0.007/100] T= Total number of policy holder of age P => T[0.88/100] If any of these policy holders (who die during the next year) is picked at random, the probability that he is a preferred policy holder is : [30*0.008*T/100] / [0.88 * T/100] = 24/88 = 3/11 = 0.2727.