A building contractor undertakes a job to construct 4

Q:- A building contractor undertakes a job to construct 4 flats on a plot along with the parking area. Due to strike the probability of many construction workers not being present 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 is 0.80. [CBSE 2023]

Let $E_1$ : Represnt the event when many workers were not present for the job;

$E_2$ : Represent the event when all workers were present, and

$E$ : Represent completing the construction work on time.

Based on the above information answer the following questions.

(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 probability that many workers are not present given that the construction work is completed on time ?

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

Solution:- Let $E_1$ : Represnt the event when many workers were not present for the job;

$E_2$ : Represent the event when all workers were present, and

E : Represent completing the construction work on time.

$P(E_1) = 0.65, P(E_2) = 1 - 0.65 = 0.35$

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

(i) The probability that all the workers are present for the job

$P(E_2) = 1 - 0.65 = 0.35$

(ii) The probability that construction will be completed on time

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

⇒ $P(E) = 0.65\times 0.35 + 0.35\times 0.80$

⇒ $P(E) = 0.2275 + 0.28 = 0.5075$

(iii) (a) The probability that many workers are not present given that the construction work is completed on time

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

⇒ $P(E_1/E) = \dfrac{0.65\times 0.35}{0.65\times 0.35 + 0.35\times 0.80}$

⇒ $P(E_1/E) = \dfrac{0.65\times 0.35}{0.35(0.65 + 0.80)}$

⇒ $P(E_1/E) = \dfrac{0.65}{1.45} = \dfrac{13}{29}$

⇒ $P(E_1/E) = 0.4482$

(iii) (b) The probability that all workers were present given that the construction job were completed on time

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

⇒ $P(E_1/E) = \dfrac{0.35\times 0.80}{0.65\times 0.35 + 0.35\times 0.80}$

⇒ $P(E_1/E) = \dfrac{0.80\times 0.35}{0.35(0.65 + 0.80)}$

⇒ $P(E_1/E) = \dfrac{0.80}{1.45} = \dfrac{16}{29}$

⇒ $P(E_1/E) = 0.5517$

Q:- 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 : [CBSE 2023]

Solution :- See full solution

Q 1:- An instructor at the astronomical centre shows three among the brightest stars in a particular constellation. Assume that the telescope is located at O(0, 0, 0) and the three stars have their locations at the points D, A and V having position vectors $2\hat{i} + 3\hat{j} + 4\hat{k}, 7\hat{i} + 5\hat{j} + 8\hat{k}$ and $-3\hat{i} + 7\hat{j} + 11\hat{k}$ respectively. [CBSE 2024]

Solution:- See full solution

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