Question

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.

Answer 1

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.