Question 1174089: Question no 2 (A): Students are randomly selected from SMIU, probability that they are from BA is ½, probability that they are from CS is 1/3 and remaining is the probability that they are from Media Studies department. Group of 12 students is selected:
(a) What is the probability that at least 6 are from BA department
(b) What is the probability that none is from media department
(c) What is the probability that at most three are from CS Department
Question No 2 (B): Assume that the number of road accidents in any month are no ofcharacters in your first name. What is the probability fewer than three accidents will occur in a particular month?
Found 2 solutions by ewatrrr, Boreal:
Answer by ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website!
Hi,
P(BA) = 1/2 and P(CS) = 1/3 and P(OTHER) = 1/6
12 Students:
(a) What is the probability that at least 6 (1 more than half) are from BA department
P (x ≥ 6BA) = 1 - P (x ≤ 5BA) = 1 - binomcdf(12, .5, 5)
(b) What is the probability that none is from media department
P (none from the media department) = 1- 1/6 = 5/6
(c) What is the probability that at most three are from CS Department
P( x ≤ 3) = binomcdf(12, .3333, 3)
__________________
Question No 2 (B): Assume that the number of road accidents in any month
are no of characters in your first name.
What is the probability fewer than three accidents will occur in a particular month?
Poisson Distribution: Am Using μ = 5
P(x ≤ 2) = poissoncdf (2,5) = .1247
Use Your calculator. Important You feel comfortable using it.
Wish You the Best in your Studies.
Answer by Boreal(15235)  (Show Source):
You can put this solution on YOUR website! 2B is a Poisson distribution with lambda= to the number of characters in your first name. P(x< =3) would be poissoncdf(char.,2)ENTER. The "2" is the largest number of accidents below 3 and includes 0 and 1 but not 3.
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p(0-5) is 0.3872, so 1-that probability is the answer or 0.6128
p(Media Dept ) is 1/6 so probability none is from median dept is (5/6)^12=0.1122
this is p (0,1,2,3):p(0) is (2/3)^12=0.0078; p(1) is 12*(2/3)^11*(1/3)=0.0424; 66*(2/3)^10*(1/9)=0.1272, the 66 coming from 12C2; and p(3)=220*(2/3)^9*(1/27)=0.2120
total is 0.3931. Also from binomialcdf(12, (1/3),3) ENTER.
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