Sum
Solve the following problem :
The probability that a lamp in the classroom will burn is 0.3. 3 lamps are fitted in the classroom. The classroom is unusable if the number of lamps burning in it is less than 2. Find the probability that the classroom cannot be used on a random occasion.
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Solution
Let X denote the number of burning lamps.
P(lamp will burn) = p = 0.3
∴ q = 1 – p = 1 – 0.3 = 0.7
Given, n = 3
∴ X ~ B(3, 0.3)
∴ The p.m.f. of X is given by
P(X = x) = `""^3"C"_x (0.3)^x (0.7)^(3 - x), x` = 0, 1, 2, 3
P(Classroom cannot be used)
= P(X < 2)
= P(X = 0 or X = 1)
= P(X = 0) + P(X = 1)
= `""^3"C"_0 (0.3)^0 (0.7)^3 + ""^3"C"_1 (0.3) (0.7)^2`
= (0.7)3 + 3 x (0.3) x (0.7)2
= 0.784
Is there an error in this question or solution?
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