XS="100%" YS="100%"/> D <ART W="50.27pt" H="25.672pt" XS="100%" YS="100%"/> F <ART W="50.27pt" H="24.387pt" XS="100%" YS="100%"/> 1.0 Thus, the probability of a student receiving a C in the course is 0.30 = 30%. The probability of a student receiving a C may also be calculated using the following equation: <ART FILE="02_14eq06.eps" H="22.047pt" XS="100%" YS="100%"/> <ART FILE="02_14eq07.eps" W="44.51pt" H="20.802pt" XS="100%" YS="100%"/>= 0.30 2-15. a. = 1/2 = 0.5 P(H) b. = = 0.5 P(T  H) P(T) c. = = (0.5)(0.5) = 0.25 P(TT) P(T) P(T) d. = = (0.5)(0.5) = 0.25 P(TH) P(T) P(H) e. We first calculate = 0.25, then calculate = (0.5)(0.5) = 0.25. To find the P(TH) P(HT) probability of one occurring, we simply add the two probabilities. The solution is 0.50. either f. one head means that we have either or Since each of these have a At least HT, TH, HH. probability of 0.25, their total probability of occurring is 0.75. On the other hand, the complement of the outcome least one head” is tails.” Thus, we could have also “at “two computed the probability from 1 = 1 0.25 = 0.75. P(TT) – – 2-16. The distribution of chips is as follows: Red 8 Green 10 White 22 = 20 a. The probability of drawing a white chip on the first draw is
XS="100%" YS="100%"/> D <ART W="50.27pt" H="25.672pt" XS="100%" YS="100%"/> F <ART W="50.27pt" H="24.387pt" XS="100%" YS="100%"/> 1.0 Thus, the probability of a student receiving a C in the course is 0.30 = 30%. The probability of a student receiving a C may also be calculated using the following equation: <ART FILE="02_14eq06.eps" H="22.047pt" XS="100%" YS="100%"/> <ART FILE="02_14eq07.eps" W="44.51pt" H="20.802pt" XS="100%" YS="100%"/>= 0.30 2-15. a. P(H) = 1/2 = 0.5 b. P(T  H) = P(T) = 0.5 c. P(TT) = P(T) P(T) = (0.5)(0.5) = 0.25 d. P(TH) = P(T) P(H) = (0.5)(0.5) = 0.25 e. We first calculate P(TH) = 0.25, then calculate P(HT) = (0.5)(0.5) = 0.25. To find the probability of either one occurring, we simply add the two probabilities. The solution is 0.50. f. At least one head means that we have either HT, TH, or HH. Since each of these have a probability of 0.25, their total probability of occurring is 0.75. On the other hand, the complement of the outcome “at least one head” is “two tails.” Thus, we could have also computed the probability from 1 – P(TT) = 1 – 0.25 = 0.75. 2-16. The distribution of chips is as follows: Red 8 Green 10 White 22 = 20 a. The probability of drawing a white chip on the first draw is
Solution-Manual-for-Quantitative-Analysis-for-Management-11E-11th-Edition.docx