# The probability that a lamp in a classroom will be burnt out is 0.3. Six such lamps are fitted in the class-room - Mathematics and Statistics

Sum

The probability that a lamp in a classroom will be burnt out is 0.3. Six such lamps are fitted in the class-room. If it is known that the classroom is unusable if the number of lamps burning in it is less than four, find the probability that the classroom cannot be used on a random occasion.

#### Solution

Let X = number of lamps burnt out in the classroom.

p = probability of a lamp in a classroom will be burnt

∴ p = 0.3 = 3/10

∴ q = 1 - p = 1 - 3/10 = 7/10

Given: n = 6

∴ X ~ B(6, 3/10)

The p.m.f. of X is given as:

P[X = x] = "^nC_x  p^x  q^(n - x)

i.e. p(x) = "^6C_x (3/10)^x (7/10)^(6 - x)

Since, the classroom is unusable if the number of lamps burning in it is less than four, therefore

P (classroom cannot be used)

= P[X < 4] = P[X = 0] + P[X = 1] + P[X = 2] + P[X = 3]

= p(0) + p(1) + p(2) + p(3)

= ""^6C_0 (3/10)^0 (7/10)^(6 - 0) + ""^6C_1 (3/10)^1 (7/10)^(6 - 1) + ""^6C_2 (3/10)^2 (7/10)^(6 - 2) + "^6C_3 (3/10)^3 (7/10)^(6 - 3)

= 1 xx 1 xx (7/10)^6 + 6(3/10) (7/10)^5 + (6 xx 5)/(1 xx 2) * (3/10)^2 (7/10)^4 + (6 xx 5 xx 4)/(1 xx 2 xx 3) * (3/10)^3 (7/10)^3

= [7^6 + 18 xx 7^5 + 15 xx 9 xx 7^4 + 20xx 27 xx 7^3] 1/10^6

= (117649 + 302526 + 324135 + 185220)/10^6

= 929530/10^6 = 0.92953

Hence, the probability that the classroom cannot be used on a random occasion is 0.92953.

#### Notes

[Note: Answer in the textbook is incorrect.]

Concept: Binomial Distribution
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#### APPEARS IN

Balbharati Mathematics and Statistics 2 (Arts and Science) 12th Standard HSC Maharashtra State Board
Chapter 8 Binomial Distribution
Miscellaneous exercise 8 | Q 6 | Page 254