Answers
At the International University, there are 100 Chinese Students, 200 Afghans, 50 Somalians, and 650 from Pakistan. Suppose ten students are chosen at random from the total of 1000 students. Calculate the probabilities for the following events:
1. All ten students are Pakistani.
2. There are exactly two Afghani students in the group of ten.
3. There are three or more Afghani students in the group.
4. The number of Afghani students is between 1 and 5, inclusive.
Solution:
Probabilities of student selected is from Pakistan= 650/1000=0.65
Probabilities of student selected is from Afghan= 200/1000=0.2
Probabilities of student selected is from China= 100/1000=0.1
Probabilities of student selected is Somalian= 50/1000=0.05
1. All ten students are Pakistani.
P (X= all ten students are Pakistani) = 10C10 (0.65)10(0.35)0 = (0.013463)(1) = 0.013463
In Excel: =BINOMDIST(10,10,0.65,FALSE)
2. There are exactly two Afghani students in the group of ten.
P(X= Exactly two students are Afghani) = 10C2 (0.2)2(0.8) = (0.04)(0.168) = 0.302
In Excel: =BINOMDIST(2,10,0.2,FALSE)
3. There are three or more Afghani students in the group.
P (X= three or more students are Afghani) = 1- P(less than three are Afghani)
= 1- [P (X=0) + P(X=1) + P(X=2)]
= 1- [ 10C0 (0.2)0(0.8)10 + 10C1 (0.2)1(0.8)9 + 10C2 (0.2)2(0.8)8 ]
= 1- [0.107+ 0.268 + 0.302]
= 1 -0.6778
=0.322
In Excel: =1- BINOMDIST(2,10,0.2,FALSE)
4. The number of Afghani students is between 1 and 5 inclusive.
P (X= Afghani students is between 1 and 5 inclusive) = 10Ci (0.2)i(0.8)10-i
=10C1 (0.2)1(0.8)10-1 + 10C2 (0.2)2(0.8)10-2 +10C3 (0.2)3(0.8)10-3 + 10C4 (0.2)4(0.8)10-4 + 10C5 (0.2)5(0.8)10-5
= 0.268 +0.301+ 0.201+ 0.088+ 0.026
= 0.886
Excel: =BINOM.DIST(5,10,0.2,TRUE)- BINOM.DIST(0,10,0.2,TRUE)