Bombing The Bridge...!!

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Probability of a bomb hitting a bridge is 1/4. If two direct hits are neede to destroy it, find the least no. of bombs required so tht the probability of the bridge being destroyed is greater than 0.9.

Am stuck here -

(4/3)ⁿ > 10/3*(n+3) ...(n=no. of trials)
Kahaan tak hit-n-trial se kaam chalega...!!!

#Mathematics #Algebra

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3 Answers

19

Debotosh.. ·2009-12-06 08:16:29

try this :
atleast (n) bombs are required so that prob of breaking the bridge is >0.9
=> nC₂ (1/4)² (3/4)^n-2 + nC₃ (1/4)³ (3/4)^n-3 + nC₄ (1/4)⁴ (3/4)^n-4 +.......nC_n (1/4)ⁿ (3/4)⁰ > 0.9

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13

Avik ·2009-12-06 08:22:22

I took it ulta....
P(bridge not being destroyed) < 0.1

ⁿC₀(3/4)ⁿ + ⁿC₁(3/4)^n-1.(1/4)ⁿ < 0.1

Solving which i get tht condition....

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19

Debotosh.. ·2009-12-06 08:27:20

i will go for some hit and trial after this !!! [6]

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