Problem 4 A car plate in the state of Minnesota consist of three digits followed

Question

Problem 4 A car plate in the state of Minnesota consist of three digits followed by three characters. Minesota (a) Find the total number of plates can be issued by the MN DMV (b) Find the number of plates who have an A at their 4th position. (c) Find the number of plates who have at least one A (d) Find the number of plates with three consecutive three digits in increasing order, i.e., 012, 123,...,789. (e) How many plates with no repeated symbols (digitsletter) does exist? () How many plates with letters in (U, V, W. x, Y, Z) does exist? Now, assume DMV issues the plate according to a uniform distribution, i.e, issuing all possible plates are equally likely (g) What is the chance of getting a plate with at least one A? (h) What is your chance to get a plate with three letters consistent with your F.M.L. (first name, middle name last name) in the right order? (i) What is the chance of seeing a plate with first digit being (strictly) less than the second one? G) A dirty car just passed by yours, and you only saw a 4 at its second digit. What is the probability that its first (lk) If someone tells you that in her car's plate the first digit is strictly less than the second one, what is the digit being (strictly) less than the second one? chance that her second digit is 4?

Explanation / Answer

(f)

For each digit we have 10 choices and for each letter we have 6 choices so possible number of plates is

10*10*10*6*6*6 = 216000

(g)

The total numebr of possible number plates is

10*10*10*26*26*26 = 17576000

The numebr of possible plates not having A is (that is each letter has 25 chocies) is

10*10*10*25*25*25 = 15625000

So possible numebr of plates having at least an A is

17576000 - 15625000 = 1951000

So the required probability is

1951000 / 17576000 = 0.111

(h)

For letters we have only one choice and for each digit we have 10 choices so required probability is

1000 / 17576000 = 0.0000569

Here 1000 shows the possible number of plates having letters in requried order.

(i)

If first digit is zero then for second digit we have 9 choices, if  first digit is 1 then for second digit we have 8 choices and so on. So possible number of choices for first and second digits is

9+8+7+6+5+4+3+2+1 = 45

And for third digit we have 10 chocies. So possible number of plates with given condition is

45 * 10 *26*26*26 = 7909200

Hence, the required probability is

7909200/ 17576000 = 0.45

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