Long Questions (Question 3 & 4)

Posted on May 17, 2020 by Myhometuition

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_{2}

(0.2)² (0.8)⁶

= 0.2936

(b)

P (X < 3)

= P (X = 0) + P (X = 1) + P (X = 2)

C_{0}

(0.2)⁰(0.8)⁸+

C_{1}

(0.2)¹(0.8)⁷ +

C_{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:

\begin{array}{l} Let X be the random variable representing the number of families \\ who own a LCD television . \\ X ~ B (n, p) \\ X ~ B (10, \frac{3}{5}) \\ p = \frac{3}{5} = 0.6 \\ q = 1 - 0.6 = 0.4 \\ P (X = r) = {}^{n}c_{r} . p^{r} . q^{n - r} \\ P (at least 8 families own a LCD television) \\ P (X \geq 8) \\ = P (X = 8) + P (X = 9) + P (X = 10) \\ = {}^{10}C_{8} {(0.6)}^{8} {(0.4)}^{2} + {}^{10}C_{9} {(0.6)}^{9} {(0.4)}^{1} + {}^{10}C_{10} {(0.6)}^{10} {(0.4)}^{0} \\ = 0.1209 + 0.0403 + 0.0060 \\ = 0.1672 \end{array}