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Worked Examples · Example 2

Q.A discrete random variable XX has the probability distribution: X=0,1,2,3X = 0,1,2,3 with P(X)=k, 2k, 3k, 4kP(X) = k,\ 2k,\ 3k,\ 4k respectively. Find the value of kk.

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✓ Free question

Step 1 — Apply the total-probability condition. For any valid pmf, ∑ip(xi)=1\sum_i p(x_i) = 1:

k+2k+3k+4k=1k + 2k + 3k + 4k = 1

10k=1  ⇒  k=110=0.110k = 1 \;\Rightarrow\; k = \dfrac{1}{10} = 0.1

Step 2 — Dual-solve check. Substitute k=0.1k=0.1 back into each probability: P(0)=0.1, P(1)=0.2, P(2)=0.3, P(3)=0.4P(0)=0.1,\ P(1)=0.2,\ P(2)=0.3,\ P(3)=0.4. Adding independently: 0.1+0.2+0.3+0.4=1.00.1+0.2+0.3+0.4 = 1.0 ✓, and every value is non-negative (since k=0.1>0k=0.1>0) ✓ — both pmf conditions hold, confirming k=0.1k=0.1 is correct.

✓Final answer

k=0.1k = 0.1, giving the distribution P(X)=0.1,0.2,0.3,0.4P(X)=0.1, 0.2, 0.3, 0.4 for X=0,1,2,3X=0,1,2,3.

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