Question

an internet search engine search for a certain key word in a sequence of independent web sites . it is believed that 20% of the web sites contain this keyword
a) compute the probability that atleast 5 of the first 10 sites contain the given key word
b) compute the probability that the search engine had to visit at least 5 sites in order to find the first occurrence of a keyword?

Answer 1

Have you tried solving it first?

Answer 2

yeah but cant understand it plz help

Answer 3

a.)1/2
b.)1/5
generally, probability of outcomes of E is just P(E)= total no. of outcomes from e/total number of outcomes I think

Answer 4

@annas hi bro.

Answer 5

hay safi wassup!

Answer 6

@annas do you have the answers for this, I think what @anonymous has posted aren't correct

Answer 7

yeah they are not correct 0.032 and 0.40

Answer 8

It's given that 20 % of the websites have the keyword
So Probability of finding the keyword on a website is \[P=20\%=\frac{20}{100}=\frac{1}
{5}\]
Probability of not finding is \[1-\frac{1}{5}=\frac{4}{5}\]

Answer 9

a) compute the probability that atleast 5 of the first 10 sites contain the given key word
So we have to find the probability of finding keyword on
5, 6, 7, 8, 9 and 10 sites
P (5) is the probability of finding keyword on 5 sites and not finding on 10
\[P(5)=(\frac{1}{5} )^5(\frac{4}{5})^5\]
similarly
\[P(6)=(\frac{1}{5} )^6(\frac{4}{5})^4\]
\[P(6)=(\frac{1}{5} )^7(\frac{4}{5})^3\]\[P(6)=(\frac{1}{5} )^8(\frac{4}{5})^2\]\[P(6)=(\frac{1}{5} )^9(\frac{4}{5})^1\]\[P(6)=(\frac{1}{5} )^{10}\]
So total probability=P(5)+P(6)+P(7)+P(8)+P(9)+P(10)

Answer 10

thanks @ash2326 really own you one for helping

Answer 11

But check the answer, I'm not getting 0.032

Answer 12

i'll check it. by the way how much you r getting

Answer 13

0.000139

Answer 14

then i think we must use some distribution

Answer 15

b) compute the probability that the search engine had to visit at least 5 sites in order to find the first occurrence of a keyword?
P=1-(P(1)+P(2)+P(3)+P(4))
P(1)= probability of finding on first website= \(\frac{1}{5}\)
P(2)=probability of finding on second website=\(\frac{4}{5} \times \frac{1}{5}\)
P(3)=probability of finding on third website=\(\frac{4}{5} \times \frac{4 }{5} \times \frac{1}{5}\)
P(4)=probability of finding on fourth website=\(\frac{4}{5} \times \frac{4 }{5} \times \frac{4 }{5} \times \frac{1}{5}\)
We have
\[P=1-(\frac{1}{5}+\frac{4}{5} \times \frac{1}{5}+\frac{4}{5} \times \frac{4}{5}\times \frac{1}{5}+\frac{4}{5} \times \frac{4}{5}\times \frac{4}{5} \times \frac{1}{5})\]

Answer 16

@ash nice buddy good going @annas in the first part atleast word is used so you have to use binomail distribution ncx p^x(q)^n-x

Answer 17

yeah i know that @deadman340