Question

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Under a specific high value of the reference mean wind velocity, the probability of glass failure of a particular type of windows at the top of a tall building is estimated at p0.01. The ncertainty stems from the fact that nominally identical panels of glass may have notably different strengths, and that wind-induced pressures on building facades are highly variable in ime and space. There are 10 top-floor windows on a building. Assuming independence among events of window breakage, determine the probability that 1)- none will break under the mean wind velocity 2)- exactly one will break under this mean wind velocity The building has n floors with 10 exposed windows each. Because of differences in design glass thickness from floor to floor and because of variation of wind properties with height, for a given reference wind velocity, the probability of any particular window breaking depends on its loor number. The probability that any particular window on floor i breaks is Pi, for i 1, 2, .., n Assuming mutual independence of the window breakage events, write an expression for the probability that at least one window will break somewhere on the exposed face of the building Comment on the adequacy of the independence assumption. If there were in fact probabilistic dependence between window breakage events, would you expect a higher or a lower value for the probability computed under (1)?

Answer 1

Solution

Let X = number of window breakages out of 10 top-floor windows. Then, X ~ B(10, p), where

p = probability of a single window breakage = 0.01 [given]………………………………….................…………… (1)

Back-up Theory

If X ~ B(n, p). i.e., X has Binomial Distribution with parameters n and p, where n = number of trials and p = probability of one success, then probability mass function (pmf) of X is given by

p(x) = P(X = x) = (ⁿC_x)(p^x)(1 - p)^{n – x}, x = 0, 1, 2, ……. , n ……………………....................…………..............…..(2)

[The above probability can also be directly obtained using Excel Function: Statistical, BINOMDIST…………….(2a)

Now to work out the solution,

Part (1)

Probability none breaks

= P(X = 0)

= 0.99¹⁰ [vide (2)]

= 0.9044 Answer

Part (2)

Probability exactly one breaks

= P(X = 1)

= 10 x 0.01 x 0.99⁹ [vide (2)]

= 0.0914 Answer

Part (3)

Given probability of a single window breakage in the ith floor = p_i, probability none breaks in ith floor

= p_i¹⁰ [vide (2)]

Probability none breaks in any of n floors

= Probability none breaks in first floor x Probability none breaks in second floor x …… x Probability none breaks in nth floor

= (p₁¹⁰) (p₂¹⁰) ……… (p_n¹⁰)

= (p_1.p_2. ……… .p_n)¹⁰ Answer

Part (4)

In all the above calculations, multiplicative law of probability could be used because we are given that window breakages are independent.

If this condition of independence is not valid, the joint probability has to be obtained by multiplying the conditional probability by the marginal probability, which will result in a lower value than in the case of independence. Answer

DONE