CA Foundation 2021 · Paper 3 · Quantitative AptitudeQ66 · 1 markVerified answer
For a probability distribution, probability is given by P(Xi)=XikP(X_i) = \dfrac{X_i}{k}; Xi=1,2,…,9X_i = 1, 2, \ldots, 9. The value of kk is :
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∑P=1⇒45/k=1⇒k=45\sum P=1 \Rightarrow 45/k=1 \Rightarrow k=45.

Step 1 — Total-probability axiom

∑i=19P(Xi)=1.\sum_{i=1}^{9} P(X_i)=1.

Step 2 — Substitute and sum

∑i=19Xik=1k∑i=19i=1k⋅9⋅102=45k.\sum_{i=1}^{9}\frac{X_i}{k}=\frac{1}{k}\sum_{i=1}^{9} i=\frac{1}{k}\cdot\frac{9\cdot 10}{2}=\frac{45}{k}.

Step 3 — Solve

45k=1⇒k=45.\frac{45}{k}=1 \Rightarrow k=45.

Watch out

The sum 1+2+⋯+91+2+\cdots+9 is 4545, not 5555 (that is 1+⋯+101+\cdots+10). Option (A) 55 is the classic off-by-one trap. …

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