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5.4.1 Probability Distributions Long Questions


Question 1:
In a school examination, 2 students out of 5 students failed Chemistry.
(a)    If 6 students are chosen at random, find the probability that not more than 2 students failed Chemistry.
(b)   If there are 200 Form 4 students in that school, find the mean and standard deviation of the number of students who failed Chemistry.
 
Solution:
(a)
X− Number of students who failed Chemistry. X~B( n,p ) X~B( 6,  2 5 ) P(X=r)= c n r . p r . q n−r P(X≤2) =P( X=0 )+P( X=1 )+P( X=2 ) = C 6 0 ( 2 5 ) 0 ( 3 5 ) 6 + C 6 1 ( 2 5 ) 1 ( 3 5 ) 5 + C 6 2 ( 2 5 ) 2 ( 3 5 ) 4 =0.0467+0.1866+0.3110 =0.5443

(b)
X~B( n,p ) X~B( 200,  2 5 ) Mean of X =np=200× 2 5 =80 Standard deviation of X = npq = 200× 2 5 × 3 5 = 48 =6.93



Question 2:
5% of the supply of mangoes received by a supermarket are rotten.
(a) If a sample of 12 mangoes is chosen at random, find the probability that at least two mangoes are rotten.
(b)  Find the minimum number of mangoes that have to be chosen so that the probability of obtaining at least one rotten mango is greater than 0.85.
 
Solution:
(a)
X ~ B (12, 0.05)
1 – P (X ≤ 1)
= 1 – [P (X = 0) + P (X = 1)]
= 1 – [  C 12 0 (0.05)0 (0.95)12 +   C 12 1 (0.05)1 (0.95)11]
= 1 – 0.8816
= 0.1184

(b)
P (X ≥ 1) > 0.85
1 – P (X = 0) > 0.85
P (X = 0) < 0.15
  C n 0 (0.05)0(0.95)n < 0.15
n lg 0.95 < lg 0.15
n > 36.98
n = 37

Therefore, the minimum number of mangoes that have to be chosen so that the probability of obtaining at least one rotten mango is greater than 0.85 is 37.



Question 3:
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)
=  C 8 2  (0.2)2 (0.8)6
= 0.2936

(b)
P (X < 3)
= P (X = 0) + P (X = 1) + P (X = 2)
= C 8 0  (0.2)0(0.8)8 +   C 8 1  (0.2)1(0.8)7 +   C 8 2  (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 families who own a LCD television. X~B( n,p ) X~B( 10,  3 5 ) p= 3 5 =0.6 q=1−0.6=0.4 P(X=r)= c n r . p r . q n−r P( at least 8 families own a LCD television ) P(X≥8) =P( X=8 )+P( X=9 )+P( X=10 ) = C 10 8 ( 0.6 ) 8 ( 0.4 ) 2 + C 10 9 ( 0.6 ) 9 ( 0.4 ) 1 + C 10 10 ( 0.6 ) 10 ( 0.4 ) 0 =0.1209+0.0403+0.0060 =0.1672

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