# 1) Calculate the mean value for the number of heads appearing in a simulations throw of 3 coins. 2) Suppose your neighbour has two children and you come to know that one of his children is a son. What is the probability that his other child is a son. 3) Assume that on a weekday one telephone number out of ever ten being called is busy. If 6 randomly selected numbers are called what is the probability that at least three of them will be busy. 4) If power failures occur according to a poisson distribution with an average of 3 failures every 20 weeks, calculate the probability that there will not be more than one failure during a particular week.

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by ItzBittuPaul

2015-04-06T00:33:45+05:30

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1)  n = number of heads in 3 throws.
total possibilities of outcomes = 2^3 = 8

n = 0,   Prob  (TTT) = 1/8
n= 1, the throws:  HTT, THT, TTH : P(n = 1) = 3/8
n= 2,  P(HHT, HTH, THH) = 3/8
n = 3 :  P(HHH) = 1/8
mean value for n :    0 *1/8 + 3/8 * 1 + 3/8 * 2 + 3 * 1/8 =
expected value for n : 12/8 = 3/2
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2)        The event of one child being a son is independent of the other child being a son or a daughter.  Hence, the probability that the other child is a son = 1/2
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3)  p = 1/10  that a called number is busy
n = number of tel. numbers busy when called.
outcome   B = busy.      F = free.

P(n=0) = P(FF FF FF) = (9/10)^6
P(n=1) =  P(PFFFFF, FPFFFF, FFPFFF, FFFPFF, FFFFPF, FFFFFP)
= 6 * (9/10)^5
P(n=2) =  1/10 * 1/10 * (9/10)^4 * ⁶C₂ = 15 /100 * (9/10)^4

P(n=3) = ⁶C₃ (1/10)^3  (9/10)^3
P(n=4) = ⁶C₄ (1/10^4 (9/10^2
P(n=5) = 6 (1/10)^5  (9/10)
P(n=6 )  =  (1/10)^6

Probability that at least three numbers called are busy
1 - P(n=0) - P(n=1)- P(n=2)
or  P(n=3) +P(n=4) + P(n=5) + P(n=6)
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Poisson's distribution.for the failures of power supply in  a week.
x = number of power supply failures in a week
expected value = avg = λ = 3/20    per week

Prob (x <= 1) =  P(x = 0) + P(x = 1)

P(x = 0 ) = e^{-3/20} = 0.86
P(x=1) =  0.129

So probability required = 0.86 + 0.129 = 0.989 ≈  0.99