A teacher lost her password to her account on the school website. She uses 4 different password for various accounts. The website allows up to three wrong attempts before banning an user. Find the probability that she will not be banned.
A teacher lost her password to her account on the school website. She uses 4 different password for various accounts. The website allows up to three wrong attempts before banning an user. Find the probability that she will not be banned.
Hi.
You need to sum all possible events like
prob("first-try password is correct") = 1/4
prob("sencond-try password is correct") = prob(first-try password was wrong)*prob("secondt-try password is correct") = 3/4 * 1/3
(there are 4 passwords, 3 of them are wrong: so 3/4 - after that try there are 3 passwords remaining, two of them are incorrect, 1 of them is correct => 1/3 )
prob("last-try password is correct") = prob("fail, fail, correct") 3/4*2/3*1/2
prob("she types in the correct password before getting banned") = 1/4+3/4*1/3 + 3/4*2/3*1/2 = 0.75 = 75% = prob("not getting banned")
Rapha
I have figured another way to complete this question. However, don't know if it's right...
$\displaystyle P(Not getting banned)=1-P(Get Banned)$
$\displaystyle =1-[\frac{3}{4} \times \frac{2}{3} \times \frac{1}{2}]$
$\displaystyle =1-(\frac{1}{4})$
$\displaystyle =\frac{3}{4}$