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Question 40 of 41
Q.

The discrete random variable X has the probability function

XX11223344
P(X=x)P(X = x)kk2k2k3k3k4k4k

Show that k=0.1k = 0.1

Puducherry TnboardTamil Nadu HSC (DGE) Commerce Board 2026Subjective· 2mImportance★★★★★
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For any probability mass function the total probability is 11. Summing k+2k+3k+4k=10kk+2k+3k+4k=10k and equating to 11 gives k=0.1k=0.1.

In the Tamil Nadu HSC Business Mathematics syllabus, a discrete probability distribution is valid only when ∑P(X=x)=1\sum P(X=x)=1 and every P(X=x)≥0P(X=x)\ge 0.

Step 1 — Add all the probabilities:

P(X=1)+P(X=2)+P(X=3)+P(X=4)=k+2k+3k+4k=10k.P(X=1)+P(X=2)+P(X=3)+P(X=4)=k+2k+3k+4k=10k.

Step 2 — Apply the total-probability condition: …

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