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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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)
=========================
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
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
==========================
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
=================
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)
=========================
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
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