Let’s assume that the last two characters of your car’s license plate number are the letter characters.
a)What’s the probability of the last two characters being AZ?
b)What’s the probability of the last character being B?
c)What’s the probability that the last digit is either A, B, or C?
Please show all work!
2006-09-22 11:38:00 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
alphabet's letters = 26 letters
+ numbers = 10 numbers
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total = 36 possibilities
P(A) = numbers of possible events / total events
1) P(A) = 36 X 36 X 36 X 36 X 36 X 1 X 1 / (36) ^ 7
= 60466176 / 78364164096
= 0.000771
2) P(B) = (36) ^ 5 X 26 X 1 / (36) ^ 7
= 1572120576 / 78364164096
= 0.020061
3) P(C) = (36) ^ 5 X 26 X 3 / (36) ^ 7
= 4716361728 / 78364164096
= 0.060185
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N.B.: 36 ^ 7 => 36 power 7 = 36 X 36 X 36 X 36 X 36 X 36 X 36
P(A) = Probability of realisation for event A
P(B) = '' '' '' '' '' B
P(C) = '' '' '' '' '' C
2006-09-22 12:47:06 · answer #1 · answered by frank 7 · 0⤊ 0⤋
a) 1 in 676
b) 1 in 26
c) 3 in 26
2006-09-22 11:41:54 · answer #2 · answered by Joe T 1 · 0⤊ 0⤋
a) 1/26 * 1/26
b) 1/26
c) 3/26
2006-09-22 11:41:45 · answer #3 · answered by dokntowhy 1 · 0⤊ 0⤋