19i will go for some hit and trial after this !!! [6]
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...!!!
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3 Answers
19try 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 +.......nCn (1/4)n (3/4)0 > 0.9
13I 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....