Question

The access code for a garage door consists of three digits. Each digit can be any number from 1 through 5​, and each digit can be repeated. Complete parts​ (a) through​ (c). ​(a) Find the number of possible access codes. ​(b) What is the probability of randomly selecting the correct access code on the first​ try? ​(c) What is the probability of not selecting the correct access code on the first​ try? ​(a) Find the number of possible access codes. The number of different codes available is nothing.

Answer 1

Answer:

(a) 125

$(b) \dfrac{1}{125}$

$(c) \dfrac{124}{125}$

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:

$5\times 5 \times 5 = 125$

(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:

$P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}$

Here, only 1 code is correct, so

Number of favorable cases = 1

Total number of cases = 125

So, required probability:

$P(E) = \dfrac{1}{125}$

(c) Probability of not selecting the correct access code on first time:

$P(\overline E) = 1-P(E)\\\Rightarrow P(\overline E) = 1-\dfrac{1}{125}\\\Rightarrow P(\overline E) = \dfrac{125-1}{125}\\\Rightarrow P(\overline E) = \dfrac{124}{125}$

So, the answers are:

(a) 125

$(b) \dfrac{1}{125}$

$(c) \dfrac{124}{125}$