x=0∑∞k(x+1)3−x=1
⇒k1=1+32+323+334+ ...(i)
3k1=31+322+333+… ...(ii)
(i)- (ii) ⇒k1−3k1=1+31+321+…
⇒k=94P(x≥3)=1−P(x=0)−P(x=1)−P(x=2)=1−k(1+32+93)=91
If the probability that the random variable X takes the value x is given by P(X=x)=k(x+1)3−x, x=0,1,2,3……, where k is a constant, then P(X≥3) is equal to
Held on 3 Apr 2025 · Verified 6 Jul 2026.
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