posted by Mackenzie
your social security number is a 9 digit number. suppose that the last 4 digits are chosen randomly from the number 0-9.
a. how many sequences are possible for the last 4 digits?
b. what is the probability of getting 5 or 0 in the last digit.
c. what is the probability of getting two 5's in the last two digits?
a.) 0000 through 9999 = 10,000 possibilities
b.) 1/10 + 1/10 = 1/5
c.) 1 (55) out of 100 (00 through 99)
for a i thought you were supposed to do 9*8*7*6 and the answer that i got was 3,024.
for b i got 9*8=72
for c i got 9*9=81
does that make sense
The numerals get smaller if you are considering probabilities without replacement. In other words, using a numeral in one position would prohibit its use in other positions. However, the use of any numeral does not effect the probabilities of the remaining numerals.
I hope this helps a little more. Thanks for asking.