Question

3.3 Counting Techniques and Probabality 81 1. Compute 8C2. 2 ! 2. Suppose 2 scholarships, $1000 each, is to be given to 2 students. There 3. Suppose 2 scholarships, one for $2000 and another for $1000, will be are 9 applicants. How many choices of 2 awardees are possible? given to 2 students. There are 9 applicants. How many choices of 2 awardees are possible? 4. In an annual sports meet, 9 students compete in 2 events (the 100 meter 0 meter race). How many ways can two winners can be picked? 5. A coin is tossed 6 times. What is the total number of possible out- and 100 comes? 6. The sex of 8 unborn children is to be determined today in a lab. What is the probability that 2 particular children (say the child of Elizabeth and the child of Maria) will be boys? 7. A committee of 8 is to be formed from a group of 7 mothers, 6 fathers, and 8 children. What is the probability that the committee will have exactly 4 fathers? 8. The 6 seats in the first row are to be assigned to a class of 1 and 20 men. What is the probability that all the seats will be to women. assigned

Answer 1

Counting Techniques and Probability 8^C_2 = 8!/2! 6! = 8 times 7 times 6!/2 times 1 times 6! = 28 Total applicants = 9 choose 2 out of 9 = 9^C_2 = 9 times 8 times 7!/2 times 7! = 36 choices 2 scholarships, 1 for $2000 & one for $1000 choose 2 out of 9 applicants = 36 ways for these 2, there are 2 choices as the first person can get either $2000 scholarship or $1000. Therefore Total ways = 36 times 2 = 72 Number of events = 2 (say A & B) 9/A 9/B then there are 9 and 9 choices as same person can win both events. Therefore Total choices = 81 coin is tossed 6 times Each time the coin is tossed, we have 2 possible outcomes. {H, T} Therefore for six tosses, 2 2 2 2 2 2 = 2^6 outcomes = 64 outcomes Total children = 8 (unborn) Number of boys possible