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Class 12 Probability · Long Answer3 Marksprobability_distributions

Class 12 Maths · Probability · Long Answer #109

X=0,1,2,3,4,5,6 has probabilities C,2C,2C,3C,C²,2C²,7C²+C. Find C and E(X).

✅ Step-by-Step Solution

Final Answer: C=1/10; E(X)=133/50=2.66
  1. Normalization gives 10C²+9C=1, or (10C−1)(C+1)=0.
  2. C=−1 gives negative probability P(X=0), so C=1/10 is the valid root.
  3. E(X)=0·C+1·2C+2·2C+3·3C+4C²+5·2C²+6(7C²+C)=21C+56C². Substitute C=1/10 to get 2.1+0.56=2.66.
  4. Thus C=1/10 and E(X)=133/50=2.66.

📋 CBSE Stepwise Marking Scheme

  • [0.5 marks] Normalization gives 10C²+9C=1, or (10C−1)(C+1)=0.
  • [1 mark] C=−1 gives negative probability P(X=0), so C=1/10 is the valid root.
  • [1 mark] E(X)=0·C+1·2C+2·2C+3·3C+4C²+5·2C²+6(7C²+C)=21C+56C². Substitute C=1/10 to get 2.1+0.56=2.66.
  • [0.5 marks] Thus C=1/10 and E(X)=133/50=2.66.

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