Answer:
There is a
probability that the correct password is found on the sixth try.
Step-by-step explanation:
We have n passwords, only 1 is correct.
So,
Since a password is put back with other, at each try, we have that:
The probability that a password is correct is ![\frac{1}{n}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bn%7D)
The probability that a password is incorrect is
.
There are 6 tries.
We want the first five to be wrong. So each one of the first five tries has a probability of
. So, for the first five tries, the probability of getting the desired outcome is
.
We want to get it right at the sixth try. The probability of sixth try being correct is
.
So, the probability that the first five tries are wrong AND the sixth is correct is:
![P = \frac{(n-1)^{5}}{n^{5}}\frac{1}{n} = \frac{(n-1)^{5}}{n^{6}}](https://tex.z-dn.net/?f=P%20%3D%20%5Cfrac%7B%28n-1%29%5E%7B5%7D%7D%7Bn%5E%7B5%7D%7D%5Cfrac%7B1%7D%7Bn%7D%20%3D%20%5Cfrac%7B%28n-1%29%5E%7B5%7D%7D%7Bn%5E%7B6%7D%7D)
There is a
probability that the correct password is found on the sixth try.
There is about a 44% chance of rolling 7
Answer:
110°
Step-by-step explanation:
mFG = 2×<FEG = 2×55 = 110°
Answered by GAUTHMATH
Hi!
To solve this problem you can set up a proportion (I like doing it this way) or you can just do:
First deal:
(24/3)*5 = 40 dollars
Second deal:
(24/4)*6 = 36 dollars
Finding money saved:
40-36 =4
They will save 4 dollars buying the second 4 for 6 dollars deal!