If x has the probability function π(π₯) = π/2^x for x=0,1,2,β¦β¦β¦ then determine the followings
a) Vale of k
b) P(xβ₯ 4)
By definition,
βx=0βk2x=1=kβx=0β12x=kβ 2\displaystyle \sum_{x=0}^\infty \frac{k}{2^x} = 1 = k\sum_{x=0}^\infty \frac{1}{2^x} = k \cdot 2x=0βββ2xkβ=1=kx=0βββ2x1β=kβ 2
k=1/2k=1/2k=1/2
P(Xβ₯4)=1βP(X<4)=1β12βx=0312x=1β12β 158=1β1516=116\displaystyle P(X \geq 4) = 1 - P(X < 4) = 1 - \frac{1}{2}\sum_{x=0}^3 \frac{1}{2^x}= 1 - \frac{1}{2} \cdot \frac{15}{8} = 1 - \frac{15}{16} = \frac{1}{16}P(Xβ₯4)=1βP(X<4)=1β21βx=0β3β2x1β=1β21ββ 815β=1β1615β=161β
Answer: k=1/2;P(Xβ₯4)=1/16.k=1/2; P(X\geq 4) = 1/16.k=1/2;P(Xβ₯4)=1/16.
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