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Question

Mathematics Question on Axiomatic Approach to Probability

Which of the following can not be a valid assignment of probabilities for outcomes of sample Space S=ω1,ω2,ω3,ω4,ω5,ω6,ω7S =\\{ω_1, ω_2 , ω_3, ω_4, ω_5, ω_6, ω_7\\}

Answer

(a)

Here, each of the numbers p(ωi) is positive and less than 1.
Sum of probabilities
=p(ω1)+p(ω2)+p(ω3)+p(ω4)+p(ω5)+p(ω6)+p(ω7)=p(ω_1)+p(ω_2)+p(ω_3)+p(ω_4)+p(ω_5)+p(ω_6)+p(ω_7)
=0.1+0.01+0.05+0.03+0.01+0.2+0.6=0.1+0.01+0.05+0.03+0.01+0.2+0.6
=1=1

Thus, the assignment is valid.

(b)

Here, each of the numbers p(ω) is positive and less than 1.
Sum of probabilities
=p(ω1)+p(ω2)+p(ω3)+p(ω4)+p(ω5)+p(ω6)+p(ω7)=p(ω_1)+p(ω_2)+p(ω_3)+p(ω_4)+p(ω_5)+p(ω_6)+p(ω_7)
=17+17+17+17+17+17+17=7×17=1=\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{1}{7}=7×\dfrac{1}{7}=1
Thus, the assignment is valid.

(c)

Here, each of the numbers p(ωi) is positive and less than 1.
Sum of probabilities
=p(ω1)+p(ω2)+p(ω3)+p(ω4)+p(ω5)+p(ω6)+p(ω7)=p(ω-1)+p(ω_2)+p(ω_3)+p(ω_4)+p(ω_5)+p(ω_6)+p(ω_7)
=0.1+0.2+0.3+0.4+0.5+0.6+0.7=0.1+0.2+0.3+0.4+0.5+0.6+0.7
=2.81=2.8≠1
Thus, the assignment is not valid.

(d)

Here, p(ω_1)$$$ and p(ω_5)$p(ω_5) are negative.
Hence, the assignment is not valid.

(e)

Here,P(ω7)=1514>1P(ω_7)=\dfrac{15}{14}>1

Hence, the assignment is not valid.