Question

a computer user tries to recall his password. she knows it can be one of 4 possible passwords. she tries her passwords until she finds the right one. let x be the number of wrong passwords she uses before she finds the right one. find E(x) and Var(x).

Answer 1

She knows it's 1 of the 4 passwords
x= no. of wrong password before she finds the right one
she can guess correct at first so x=0
Probability of this
\[P(x=0)= \frac{1}{4}\]
she guesses correct at second time
\[P(x=1)= P(getting\ wrong\ at\ first) \times P( getting\ right\ at\ second\ time)\]
\[P(x=1)=\frac{3}{4} \times \frac{1}{3}\]

Answer 2

Did you understand till this??
then we'll proceed

Answer 3

what is math processing error...??

Answer 4

Oh just reload the page

Answer 5

understand till this.... you can procced ...?

Answer 6

Just wait for sometime. I'll solve it

Answer 7

So we have
\[P(x=1)= \frac{1}{4}\]
Now Let's find the probability of getting two wrong before getting right
\[P(x=2)= \frac{3}{4} \frac{2}{3} \frac{1}{2}=\frac{1}{4}\]
Probability of getting three wrongs
\[P(x=3)= \frac{3}{4} \frac{2}{3} \frac{1}{2} \frac{1}{1}=\frac{1}{4}\]
So we have
x P(x)
0 1/4
1 1/4
2 1/4
3 1/4
\[E(x)=\sum_{x=0}^{x=3} x P(x)= 0 * 1/4+1*1/4+2*1/4+3*1/4=1.5\]

Answer 8

Did you understand this?

Answer 9

yes..... i do understand

Answer 10

Great, now can you find variance?

Answer 11

yes..

Answer 12

Good :)