Answer:
(a) 125
[tex](b) \dfrac{1}{125}[/tex]
[tex](c) \dfrac{124}{125}[/tex]
Step-by-step explanation:
We are given that access code consists of 3 digits.
Each digit can be any digit through 1 to 5 and can be repeated.
Now, this problem is equivalent to the problem that we have to find:
The number of 3 digit numbers that can be formed using the digits 1 to 5 with repetition allowed.
(a) We have 3 places here, unit's, ten's and hundred's places respectively and each of the 3 places have 5 possibilities (any digit allowed with repetition).
So, total number of access codes possible:
[tex]5\times 5 \times 5 = 125[/tex]
(b) Suppose, an access code is randomly selected, what is the probability that it will be correct.
Formula for probability of an event E can be observed as:
[tex]P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}[/tex]
Here, only 1 code is correct, so
Number of favorable cases = 1
Total number of cases = 125
So, required probability:
[tex]P(E) = \dfrac{1}{125}[/tex]
(c) Probability of not selecting the correct access code on first time:
[tex]P(\overline E) = 1-P(E)\\\Rightarrow P(\overline E) = 1-\dfrac{1}{125}\\\Rightarrow P(\overline E) = \dfrac{125-1}{125}\\\Rightarrow P(\overline E) = \dfrac{124}{125}[/tex]
So, the answers are:
(a) 125
[tex](b) \dfrac{1}{125}[/tex]
[tex](c) \dfrac{124}{125}[/tex]
