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Question

i have done it like this.....

(1/8)3+ 7/(8)3=8-2    .....how this is wrong...???

Comments

saxena

by  using 4C3 ...means u have considered that all the element all different...but may be that is not the case....there may be a hash function which give different slot for different element....

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@saxena your method is correct but just one correction is,

When we choose any 3 out of 4, then prob of 1st one is 100% (1st can go into any slot)
Hence 4C3∗(8/8)*(1/8)*(1/8)∗7/8+4C4∗(1/8)^3=29∗8^-3

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 Ashwin Kulkarni Check it!

Answer 1

Case I : All going to one slot

$(\frac{1}{8})^3  \ 1^{st} insertion=1 \  rest = \frac{1}{8}$

Case 2 :Second third or fourth  any one is going to different slot :  

 $ \frac{21}{8^3} \ 1^{st} insertion=1 \ $ then $3_{c_2}*(\frac{1}{8})^2 * \frac{7}{8}$

Case 3 : 1st one is going different and other three insertions are going to one slot :

$ \frac{7}{8^3} \ 1^{st} insertion=1 \ $ then $ \frac{7}{8}*(\frac{1}{8})^2$

Adding all of these : $\frac{1}{8^3} + \frac{21}{8^3} + \frac{7}{8^3}= \frac{29}{8^3}$

Comments

Now it is correct :)