probability q

2 Answers

ans > option (a)
ⁿC₀(1-p)ⁿ + ⁿC₁(1-p)^n-1 p +.......ⁿC_n(1-p)pⁿ =1

=> prob of atleast one success = (terms in bold letters) = 1- ⁿC₀(1-p)ⁿ â‰¥9/10
....solving we get n> 1/ {log₁₀4 - log ₁₀3 }

thanks debo :)

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