SUDHANSHUKNIT99 wrote:
jnelson5446 wrote:
dzyubam says the answer is (235/250)^5
GMAT TIGER says the answer is (235c10 / 250c10)^5
Which is correct?
i think ..
(235/250)^5----this is the probability to select any non infected computer consecutively 5 time
(235c10 / 250c10)^5----this is the probability to select 10 non infected computer consecutively 5 time
Based on the above, we see that:
P(Picking first uninfected computer) = 235/250
P(Picking second uninfected computer) = (235 - 1)/(250 - 1) = 234/249
P(Picking third uninfected computer) = (234 - 1)/(249 - 1) = 233/248
...
P(Picking 10th uninfected computer) = (227 - 1)/(242 - 1) = 226/241
In order to figure out the probability for a day's worth of scanning and not picking/finding an infected machine, we multiply the series of probabilities listed above = P(scanning 10 and finding nothing)
In order to repeat the above process for 5 days = P(scanning 10 and finding nothing)^5.
This seems like a very complex calculation, but with a bit of computational magic, we can verify that the this solution matches "(235c10 / 250c10)^5".
((235/250) (234/249) (233/248) (232/247) (231/246) (230/245) (229/244) (228/243) (227/242) (226/241))^5 - (Binomial[235, 10]/Binomial[250, 10])^5
The above equation nets out to zero! I hope this helps.