Question

 In a school, 30 percent of the students do not travel by bus. From a sample

   of 10 students find the probability that (a) at least 3 students travel by bus

   (b) at most 4 students travel by bus (c) between 2 and 7 clients students

   travel by bus 

Answer 1

Probability of people travel by bus p=(100-30)%=70%=0.7

Probability of people do not travel by bus q=30%=0.3

Here n=10,

(a) Probability at least 3 students travel by bus-

$P(X\ge 3)=1-P(X<3)$

$=1-[P(X=0)+P(X=1)+P(X=2)]\\=1-[^{10}C_0p^0q^{10}+^{10}C_1p^1q^9+^{10}C_2p^2q^8]\\=1-[(0.3)^{10}+10(0.7)(0.3)^9+45(0.7)^2(0.3)^8]\\=1-(0.00000590+0.000137+0.001446)\\=1-0.00158=0.99841$

(b) Probability at most 4 students travel by bus-

$P(X\le 4)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)$

$=^{10}C_0p^0q^{10}+^{10}C_1p^1q^9+^{10}C_2p^2q^8+^{10}C_3p^3q^7+^{10}C_4p^4q^6\\=(0.3)^{10}+10(0.7)(0.3)^9+45(0.7)^2(0.3)^8+120(0.7)^3(0.3)^7+210(0.7)^4(0.3)^6$

$=0.0000059+0.000137+0.001446+0.009+0.03675\\=0.047345$

(C) Probability that  between 2 and 7 clients students

   travel by bus -

P(2<X<7)=P(X=3)+P(X=4)+P(X=5)+P(X=6)

$=^{10}C_3p^3q^7+^{10}C_4p^4q^6+^{10}C_5p^5q^5+^{10}C_6p^6q^4\\=0.009+0.03675+0.1029+0.11435 \\=0.263$