Class 12 Probability · Long Answer3 Marksprobability_distributions
Class 12 Maths · Probability · Long Answer #62
X=0,1,2,3,4,5,6,7 has probabilities 0,k,2k,2k,3k,k²,2k²,7k²+k. Find (i) k; (ii) P(X<3); (iii) P(X>6); (iv) P(0<X<3).
✅ Step-by-Step Solution
Final Answer: k=1/10; P(X<3)=3/10; P(X>6)=17/100; P(0<X<3)=3/10
- Normalization gives 10k²+9k=1, or (10k−1)(k+1)=0.
- The roots are k=1/10,−1. The root −1 makes P(X=1) negative, so only k=1/10 is admissible.
- P(X<3)=0+k+2k=3/10; P(X>6)=7k²+k=17/100; P(0<X<3)=k+2k=3/10.
- Therefore k=1/10 and the three probabilities are 3/10,17/100,3/10.
📋 CBSE Stepwise Marking Scheme
- [0.5 marks] Normalization gives 10k²+9k=1, or (10k−1)(k+1)=0.
- [1 mark] The roots are k=1/10,−1. The root −1 makes P(X=1) negative, so only k=1/10 is admissible.
- [1 mark] P(X<3)=0+k+2k=3/10; P(X>6)=7k²+k=17/100; P(0<X<3)=k+2k=3/10.
- [0.5 marks] Therefore k=1/10 and the three probabilities are 3/10,17/100,3/10.
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