In a certain bowl of M&M; candies, there are 10 yellow plain ones, 6 orange pean
ID: 3324945 • Letter: I
Question
In a certain bowl of M&M; candies, there are 10 yellow plain ones, 6 orange peanut ones, 6 yellow peanut ones, 5 green peanut ones, 3 orange plain ones and 5 green plain ones. What is the probability that: a) the selected candy will be green? b) the selected candy will be plain? c) the selected candy will be both green and plain? d) the selected candy will be either green or plain? e) the selected candy will be green, given that it is plain? the selected candy will be plain, given that it is green? g) If someone selects two candies from the bowl, that they will both be orange? h) If someone selects three candies from the bowl,that that at least one of them is yellow? 6) Suppose that 13% of people are left hand d and that 6 people are selected at random, a) What is the probability that exactly 2 of them are left handed? b) What is the probability that 5 or more are left handed? What is the mean and standard deviation of the number of left handed people out of the 6 chosen? c) d) Would it be unusual to have all of the selected people be left handed? A multiple choice test has9 questions each of which has 5 possible answers, only ore of which iscorrect.If Judy, who forgot to stucy for the test, gusses on all questions what is the pradbability that she willanswer exactly 3 questiors correctly?Explanation / Answer
5)a) probability = 10/35 = 0.2857
B) probability = 18/35 = 0.524
C) probability = 5/35 = 0.143
D) probability = P(g) + P(P) - P(g and P) = 10/35 + 18/35 - 5/35 = 0.657
E) P(g|p) = P(g and P)/P(P) = 5/18 = 0.278
F) P(P|g) = P(P and g)/P(g) = 5/10 = 0.5
G) probability = 9C2/35C2 = 0.061
H) probability = (16C1 * 19C2 + 16C2 * 19C1 + 16C3)/35C3 = 0.852
6) P = 0.13
n = 6
P(X = x) = nCx * Px * (1 - P)n - x
A) P(x = 2) = 6C2 * (0.13)^2 * (0.87)^4 = 0.145
B) P(x > 5) = P (x = 5) + P(x = 6)
= 6C5 * (0.13)^5 * (0.87)^1 + 6C6 * (0.13)^6 * (0.87)^0 = 0.0002
C) mean = n * P = 6 * 0.13 = 0.78
Standard deviation = sqrt(n * P * (1 - P)) = sqrt(6 * 0.13 * 0.87) = 0.824
D) P(x = 6) = 6C6 * (0.13)^6 * (0.87)^0 = 0.0000048
7) P = 1/5 = 0.2
n = 9
P(x = 3) = 9C3 * (0.2)^3 * (0.8)^6 = 0.176