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slava [35]
4 years ago
14

License plate numbers in a certain state consists of seven characters. The first character is a non-zero digit (1 through 9). Th

e next four characters are capital letters (A through Z) and the last two characters are non-zero digits. Therefore, a license plate number in this state can be any string of the form:
Digit-Letter-Letter-Letter-Letter-Digit-Digit

(a) How many different liscence plate numbers are possible?
(b) How man licence plate numbers are possible if no digit appears more than once?
(c) How man licence plate numbers are possible if no digit or letter appears more than once?
Mathematics
1 answer:
xeze [42]4 years ago
6 0

Answer:

a)  333,135,504 different plates

b) 230,315,904 different plates

c) 180,835,200 different plates

Step-by-step explanation:

Pattern: Digit(1-9)-Letter-Letter-Letter-Letter-Digit(1-9)-Digit (1-9)

We will calculate the number of possibilities for the digits part, then for the letters part, then we'll multiply them together.

For the digits, we have 3 numbers, first and last 2 positions. We can consider this is a single 3-digit number, where n = 9 (since they are non-zero digits) and r = 3.  

For the letters part, it's basically a 4-letter word, where n = 26 (A through Z) and r = 4.

(a) How many different license plate numbers are possible?

No limitation on repeats for this question:

For the digits, we have 9 * 9 * 9 = 729 (since repetition is allowed, and we can pick any digit from 0 to 9 for each position)

For the letters we have: 26 * 26 * 26 * 26 = 456,976

Because the digits and letters arrangements are independent from each other, we multiply the two numbers of possibilities to have the global number of possibilities:

P = 729 * 456976 = 333,135,504 different plates, when there's no repeat limitation.

(b) How man license plate numbers are possible if no digit appears more than once?

Repeats limitation on digits:

For the digits, we have 9 * 8 * 7 = 504 (since repetition is NOT allowed, we can pick any of 9 digits for first position, then any 8 remaining and finally any 7 remaining at the end)

For the letters we still have: 26 * 26 * 26 * 26 = 456,976

Because the digits and letters arrangements are independent from each other, we multiply the two numbers of possibilities to have the global number of possibilities:

P = 504 * 456976 = 230,315,904 different plates, when there's no repeat on the digits.

(c) How man license plate numbers are possible if no digit or letter appears more than once?

Repeats limitation on both digits and letters:

For the digits, we have 9 * 8 * 7 = 504 (

For the letters we still have: 26 * 25 * 24 * 23 = 358,800

Because the digits and letters arrangements are independent from each other, we multiply the two numbers of possibilities to have the global number of possibilities:

P = 504 * 358800 = 180,835,200 different plates, when there's no repeat on the digits AND on the letters.

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We find fraction of whole wheat bagels to the sample for each shop using given information

Shop A = \frac{10}{50}  = 0.20

Shop B = \frac{23}{100}  = 0.23

Shop C = \frac{7}{25}  = 0.28

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Shop A = 0.20 * 500= 100

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Answer:

There is a 21.053% probability that this person made a day visit.

There is a 39.474% probability that this person made a one night visit.

There is a 39.474% probability that this person made a two night visit.

Step-by-step explanation:

We have these following percentages

20% select a day visit

50% select a one-night visit

30% select a two-night visit

40% of the day visitors make a purchase

30% of one night visitors make a purchase

50% of two night visitors make a purchase

The first step to solve this problem is finding the probability that a randomly selected visitor makes a purchase. So:

P = 0.2(0.4) + 0.5(0.3) + 0.3(0.5) = 0.38

There is a 38% probability that a randomly selected visitor makes a purchase.

Now, as for the questions, we can formulate them as the following problem:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

P = \frac{P(B).P(A/B)}{P(A)}

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

Suppose a visitor is randomly selected and is found to have made a purchase.

How likely is it that this person made a day visit?

What is the probability that this person made a day visit, given that she made a purchase?

P(B) is the probability that the person made a day visit. So P(B) = 0.20

P(A/B) is the probability that the person who made a day visit made a purchase. So P(A/B) = 0.4

P(A) is the probability that the person made a purchase. So P(A) = 0.38

So

P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.4*0.2}{0.38} = 0.21053

There is a 21.053% probability that this person made a day visit.

How likely is it that this person made a one-night visit?

What is the probability that this person made a one night visit, given that she made a purchase?

P(B) is the probability that the person made a one night visit. So P(B) = 0.50

P(A/B) is the probability that the person who made a one night visit made a purchase. So P(A/B) = 0.3

P(A) is the probability that the person made a purchase. So P(A) = 0.38

So

P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.5*0.3}{0.38} = 0.39474

There is a 39.474% probability that this person made a one night visit.

How likely is it that this person made a two-night visit?

What is the probability that this person made a two night visit, given that she made a purchase?

P(B) is the probability that the person made a two night visit. So P(B) = 0.30

P(A/B) is the probability that the person who made a two night visit made a purchase. So P(A/B) = 0.5

P(A) is the probability that the person made a purchase. So P(A) = 0.38

So

P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.3*0.5}{0.38} = 0.39474

There is a 39.474% probability that this person made a two night visit.

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4 years ago
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