Question 3:
Solution:
(a)
(b)
The result of a study shows that 20% of students failed the Form 5 examination in a school. If 8 students from the school are chosen at random, calculate the probability that
(a) exactly 2 of them who failed,
(b) less than 3 of them who failed.
Solution:
(a)
p = 20% = 0.2,
q = 1 – 0.2 = 0.8
X ~ B (8, 0.2)
P (X = 2)
=
C82
(0.2)2 (0.8)6
= 0.2936
(b)
P (X < 3)
= P (X = 0) + P (X = 1) + P (X = 2)
=
C80
(0.2)0(0.8)8 +
C81
(0.2)1(0.8)7 +
C82
(0.2)2(0.8)6
= 0.16777 + 0.33554 + 0.29360
= 0.79691
Question 4:
In a survey carried out in a particular district, it is found that three out of five families own a LCD television.
If 10 families are chosen at random from the district, calculate the probability that at least 8 families own a LCD television.
Solution:
Let X be the random variable representing the number of familieswho own a LCD television.X~B(n,p)X~B(10, 35)p=35=0.6q=1−0.6=0.4P(X=r)=cnr.pr.qn−rP(at least 8 families own a LCD television)P(X≥8)=P(X=8)+P(X=9)+P(X=10)=C108(0.6)8(0.4)2+C109(0.6)9(0.4)1+C1010(0.6)10(0.4)0=0.1209+0.0403+0.0060=0.1672
In a survey carried out in a particular district, it is found that three out of five families own a LCD television.
If 10 families are chosen at random from the district, calculate the probability that at least 8 families own a LCD television.
Solution:
Let X be the random variable representing the number of familieswho own a LCD television.X~B(n,p)X~B(10, 35)p=35=0.6q=1−0.6=0.4P(X=r)=cnr.pr.qn−rP(at least 8 families own a LCD television)P(X≥8)=P(X=8)+P(X=9)+P(X=10)=C108(0.6)8(0.4)2+C109(0.6)9(0.4)1+C1010(0.6)10(0.4)0=0.1209+0.0403+0.0060=0.1672