They will see each other every 30 days but won't every 15 days
(a) there are 8C2 = 28 ways of picking 2 girls from 8
And there are 21C4 = 5985 ways of picking 4 boys
Required number of ways for 2g / 4b = 28 * 5985 = 167,580
(b) at least 2 girls means combinations of 2g/4b , 3g,3b , 4g/2b , 5g 1b or
6 girls.
2g/4b = 167,580 ways
3g/3b = 8C3 * 21C3 = 56 * 1330 = 74,480
4g/2b = 8C4* 21C2 = 70 * 210 = 14,700
5g 1b = 8C5* 21 = 56*21 = 1176
6 girls = 8C6 = 28
adding these up we get the answer to (b) which is 257,964
Answer:
5 is the mean
Step-by-step explanation:
Answer:
0.0032
The complete question as seen in other website:
There are 111 students in a nutrition class. The instructor must choose two students at random Students in a Nutrition Class Nutrition majors Academic Year Freshmen non-Nutrition majors 17 18 Sophomores Juniors 13 Seniors 18 Copy Data. What is the probability that a senior Nutrition major and then a junior Nutrition major are chosen at random? Express your answer as a fraction or a decimal number rounded to four decimal places.
Step-by-step explanation:
Total number of in a nutrition class = 111 students
To determine the probability that the two students chosen at random is a junior non-Nutrition major and then a sophomore Nutrition major, we would find the probability of each of them.
Let the probability of choosing a junior non-Nutrition major = Pr (j non-N)
Pr (j non-N) = (number of junior non-Nutrition major)/(total number students in nutrition class)
There are 13 number of junior non-Nutrition major
Pr (j non-N) = 13/111
Let the probability of choosing a sophomore Nutrition major = Pr (S N-major)
Pr (S N-major)= (number of sophomore Nutrition major)/(total number students in nutrition class)
There are 3 number of sophomore Nutrition major
Pr (S N-major) = 3/111
The probability that the two students chosen at random is a junior non-Nutrition major and then a sophomore Nutrition major = 13/111 × 3/111
= 39/12321
= 0.0032