1

(a)How many license plates are possible if repetition of symbols is allowed?

I separated the two equations to show the two different cases (three or four of the digits case)

  • This is what I got: (26^3)(10^3) + (26^3)(10^4)

(b)How many license plates do not contain any repeated symbol?

  • This is what I got: (26*25*24*10*9*8) + (26*25*24*10*9*8*7)

(c)How many license plates have at least one repeated symbol?

  • This is what I got: (26*25*24*10*9*8) + (26*25*24*10*9*8*7) + (26*26*26*10*9*8) + (26*26*26*10*9*8*7)

(d)What is the probability that a license plate chosen at random has a repeated symbol?

  • I didn't know how to do this one, but i assume, if my part c is right, that to get the answer, you divide part c by part a

If you guys can check my work and see if I approached the question right will be greatly appreciated.

Blue Lightning
  • 29

1 Answers1

0

A) How many total.

You did this exactly right.

B) How many with no repeats.

You did this exactly right.

C) Wow many with at least one repeat.

I do not understand why you wrote what you wrote.

But notice #with no repeats + #with at leas one repeat = # of total possible.

So $ (26*25*24*10*9*8) + (26*25*24*10*9*8*7) + $ # with at least one repeat = $ (26^3)(10^3) + (26^3)(10^4)$ or

# with at least one repeat $= (26^3)(10^3) + (26^3)(10^4)- (26*25*24*10*9*8) + (26*25*24*10*9*8*7)$

D) $P = \frac {\#desired outcomes}{\# total outcomes}=\frac CA = \frac{(26^3)(10^3) + (26^3)(10^4)- (26*25*24*10*9*8) + (26*25*24*10*9*8*7)}{(26^3)(10^3) + (26^3)(10^4)} = 1 - \frac{(26*25*24*10*9*8) + (26*25*24*10*9*8*7)}{(26^3)(10^3) + (26^3)(10^4)} = 1 - P(\text{no repeated characters}) $

Worth noting: If two events $A$ and $B$ are mutually exclusive and exhaustive, then $P(A) = 1 - P(B)$.

====

BTW: $\frac{(26^3)(10^3) + (26^3)(10^4)- (26*25*24*10*9*8) + (26*25*24*10*9*8*7)}{(26^3)(10^3) + (26^3)(10^4)} = 1 - \frac{(26*25*24*10*9*8) + (26*25*24*10*9*8*7)}{(26^3)(10^3) + (26^3)(10^4)}=$

$1 - \frac{(26*25*24*10*9*8)(1+ 7)}{26^3*11000}=$

$1 - \frac {24*9*8*8}{26^2*44}$

$1 - \frac {2^{9}*3^3}{2^4*11*13^2}= 1- \frac{2^5*27}{11*13^2} = 1-\frac{864}{1859} $

....Not that it really matters.

fleablood
  • 124,253