Answer:
A. 0.22
B. 0.18
C. 0.25
D. 0.244
Step-by-step explanation:
S = {51 to 100} = 50
The sample space S contains values from 51 to 100 which is a total of 50 different values.
A.
Probability of A (lies between the values of 90 to 100 = 11).
11/50 = 0.22
B.
For a student to fail the course, his course has to be less than 60 = from 51 to 59. A total of 9 values.
9/50 = 0.18
C.
For student to get c, (70 to 79) a total of 10 values: 10/50 = 0.20
P(student did not get C) = 1-0.20 = 0.80
To get B, ( 80 to 89)
10/50 = 0.20
Probability that a student who is known not to have a c grade has a b grade = 0.20/0.80 = 0.25
D.
Probability of passing lies between 60 to 100 = 41 scores
41/50 = 0.82
Probability of student who passed having a B = 0.20/0.82 = 0.244
Answer:
b. 8 is the correct answet it's clearly
your answer for this question is y=c hope this helps
The order does not matter, so we will be using combinations here. 52C1 is the first one since there is only one possibility. 16C1 is the second. You have to multiply these two, so 52 X 16 = 832.
You would have to do 1/3x2/5 which is 2/15. So 2/15 of the girls in the school play sports.