(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.
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 $P(A) = 1 - P(B)$.
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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.