Question

A five​-digit number starts with a number between 4​-9 in the first​ position, with no restrictions on the remaining 4 digits. a) Find the probability that a​ randomly-chosen phone number contains all different digits. b) Find the probability that a​ randomly-chosen phone number contains at least one repeated digit. ​a) Write an expression that models the probability. Select the correct choice below and fill in the answer​ box(es) within your choice.

Answer 1

Answer:

(a) $Pr = 0.3024$

(b) $Pr = 0.6976$

(c) $Pr = \frac{^9P_{n-1}}{10^{n-1}}$

Step-by-step explanation:

Given

$Start = \{5,6,7,8\}$ i.e. between 4 and 9

$n(Start) =4$

$Digits = 5$

Solving (a): Probability that each of the 5 digit are different

Since there is no restriction;

The total possible selection is as follows:

$First\ digit = 4$ (i.e. any of the 4 start digits)

$Second\ digit = 10$ (i.e. any of the 10 digits 0 - 9)

$Third\ digit = 10$ (i.e. any of the 10 digits 0 - 9)

$Fourth\ digit = 10$ (i.e. any of the 10 digits 0 - 9)

$Fifth\ digit = 10$ (i.e. any of the 10 digits 0 - 9)

So, the total is:

$Total = 4 * 10 * 10 * 10 * 10$

$Total = 40000$

For selection that all digits are different, the selection is:

$First\ digit = 4$ (i.e. any of the 4 start digits)

$Second\ digit = 9$ (i.e. any of the remaining 9)

$Third\ digit = 8$ (i.e. any of the remaining 8)

$Fourth\ digit = 7$ (i.e. any of the remaining 7)

$Fifth\ digit = 6$ (i.e. any of the remaining 6)

So:

$Selection =4 * 9 * 8 * 7 * 6$

$Selection =12096$

So, the probability is:

$Pr = \frac{Selection}{Total}$

$Pr = \frac{12096}{40000}$

$Pr = 0.3024$

Solving (b): At least 1 repeated digit

The probability calculated in (a) is the all digits are different i.e. P(None)

So, using laws of complement

We have:

$P(At\ least\ 1) = 1 - P(None)$

So, we have:

$Pr= 1 - 0.3024$

$Pr = 0.6976$

Solving (c): An expression to model the probability.

Using (a) as a point of reference, we have;

$Pr = \frac{Selection}{Total}$

Where

$Selection =4 * 9 * 8 * 7 * 6$ ---- for selection of 5 i.e. n = 5

$Total = 4 * 10 * 10 * 10 * 10$

$Selection =4 * 9 * 8 * 7 * 6$

This can be rewritten as:

$Selection = 4 * ^9P_4$

4 can be expressed as: 5 - 1

So, we have:

$Selection = (5-1) *^9P_{5-1}$

Substitute n for 5

$Selection = (n-1) *^9P_{n-1}$

$Selection = (n-1)^9P_{n-1}$

$Total = 4 * 10 * 10 * 10 * 10$

This can be rewritten as:

$Total = 4 * 10^4$

$Total = (5-1) * 10^{5-1}$

$Total = (n-1) * 10^{n-1}$

$Total = (n-1) 10^{n-1}$

So, the expression is:

$Pr = \frac{(n-1)^9P_{n-1}}{(n-1)10^{n-1}}$

$Pr = \frac{^9P_{n-1}}{10^{n-1}}$

Where n represents the digit number