A building contractor undertakes a job to construct 4

Case study 1:

A building contractor undertakes a job to construct 4 flats on a plot along with parking area. Due to strike the probability of many construction workers not being present for the job is 0.65. The probability that many are not present and still the work get completed on times is 0.35. The probability that work will be completed on time when all workers are present 0.80.   

Let: E_1: reprsent the event many workers were not present for the job;

E_2: represent the event when all workrs were present ; and

E: represent completing the constuction work on time

Based on the above information answer the following question

(i) What is the probability that all the workers are present for the job ?

(ii) What is the probability that construction will be completed on time ?

(iii) (a) What is the probaility that many worker are not present given that the construction work is completed on time ?

OR

(iii) (b) What is the probability that all workers were present given that the construction job was completed on time.

[CBSE  2023]

Solution:

(i) Due to strike the probability of many construction workers not being present for the job is 0.65

E_2 = represent the event when all workrs were present

Hence the probability of all workers are present

P(E_2) = 1 – 0.65 = 0.35

(ii) Given that,

P(E_1) = 0.65, P(E_2) = 0.35

P(E/E_1) = 0.35, P(E/E_2) = 0.80

The probability that construction will be completed on time

P(E) = P(E_1)P(E/E_1) + P(E_2).P(E/E_2)

=  0.65 × 0.35 + 0.35 × 0.80

=  0.2275 + 0.28

= 0.5075

(iii) (a) Given that,

P(E_1) = 0.65, P(E_2) = 0.35

P(E/E_1) = 0.35, P(E/E_2) = 0.80

The probaility that many worker are not present given that the construction work is completed on time

P(E_1/E) = \dfrac{P(E_1)P(E/E_1)}{P(E_1)P(E/E_1) + P(E_2).P(E/E_2)}

= \dfrac{0.65 \times 0.35}{ 0.65 \times 0.35 + 0.35 \times 0.80}

= \dfrac{0.2275}{0.2275 + 0.28}

= \dfrac{0.2275}{0.5075}

= 0.4482 = 0.45

OR

(iii) (b)Given that,

P(E_1) = 0.65, P(E_2) = 0.35

P(E/E_1) = 0.35, P(E/E_2) = 0.80

The probability that all workers were present given that the construction job was completed on time

P(E_2/E) = \dfrac{P(E_2)P(E/E_2)}{P(E_1)P(E/E_1) + P(E_2).P(E/E_2)}

= \dfrac{0.35 \times 0.80}{ 0.65 \times 0.35 + 0.35 \times 0.80}

= \dfrac{0.28}{0.2275 + 0.28}

= \dfrac{0.28}{0.5075}

= 0.55


Case study 2:

Sooraj’s father wants to construct a rectangular garden using a brick wall on one side of the garden and wire fencing for the other three sides as shown in the figure. He has 200 metres of fencing wire.

Based on the above information, answer the following question.

(i) Let ‘x’ metres denote the length of the side of the garden perpendicular to the brick wall and ‘y’ metres denote the length of the side parallel to the brick wall.  Determine the relation representing the total length of fencing wire and also write A(x), the area of garden.

(ii) Determine the maximum value of A(x). 

Solution:

(i) Length of the rectangular field = x m

Breadth of the rectangular field = y m

Length of the fencing = 200 m

Total fencing = x + x + y

⇒  200 = 2x + y

⇒  y = 200 – 2x

Area of the garden (A) = x × y

⇒ A(x) = x(200 – 2x)

⇒ A(x) = 200 x – 2x²

(ii) Area of the recangular field

A(x) = 200 x – 2x²

Differentiate with respect to x

\frac{d}{dx}A(x) = 200 - 4x

For max and minima \frac{d}{dx}A(x) = 0

200 – 4x = 0

⇒ x  = 50

Again differentiate with respect to x

\dfrac{d^2}{dx}A(x) = -4 < 0

Hence, the area of rectangle is max when x = 50

Maximum area A(x) = 200(50) – 2(50)²

=  10000 – 5000

=  5000 m²


 

Case Study

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