NCERT Class 12 · Ch 13 Miscellaneous Exercise · Q4
NCERT Class 12 Maths · Chapter 13 Probability · Miscellaneous Exercise Question 4
90% of people are right-handed. For 10 independent random people, find P(at most 6 right-handed).
💡 Concept Hints
- Define the events and identify which event is being conditioned on.
- Use P(E|F)=P(E∩F)/P(F), or separate the possible sources before applying total probability and Bayes.
✅ Step-by-Step Solution
Final Answer: 7996999/625000000 ≈ 0.0127951984000
- Let X be the count; under independent sampling with common probability 0.9, X~Binomial(10,0.9).
- P(X≤6)=Σₖ₌₀⁶ choose(10,k)(0.9)ᵏ(0.1)^(10−k). Evaluating gives 7996999/625000000.
- Equivalently subtract the k=7,8,9,10 terms from 1; the two calculations agree.
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