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Question 27 of 41
Q.

The following information is the probability distribution of successes.

No. of successes X=xX=x012
Probability P(x)P(x)611\dfrac{6}{11}922\dfrac{9}{22}122\dfrac{1}{22}

Determine the expected number of success.

Tamil Nadu DgeTamil Nadu HSC (DGE) Commerce Board 2023Subjective· 2mImportance★★★★★
66% · 27/41 Questions
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The expected number of successes is E(X)=∑xP(x)=12E(X)=\sum xP(x)=\dfrac12.

In this Tamil Nadu HSC Business Maths & Statistics problem XX takes values 0,1,20,1,2 with probabilities 611,922,122\dfrac{6}{11},\dfrac{9}{22},\dfrac{1}{22}. The expectation is the probability-weighted mean:

E(X)=∑x P(x).E(X)=\sum x\,P(x).

E(X)=0⋅611+1⋅922+2⋅122=0+922+222=1122=12.E(X)=0\cdot\frac{6}{11}+1\cdot\frac{9}{22}+2\cdot\frac{1}{22}=0+\frac{9}{22}+\frac{2}{22}=\frac{11}{22}=\frac12.

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