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.
To answer these questions, we'll break down the problem step by step.
a. The last four digits of a social security number are chosen randomly from the numbers 0-9. Since each digit can be any number from 0 to 9, there are 10 possible choices for each of the four digits. Therefore, the total number of sequences possible for the last four digits is calculated by multiplying the number of choices for each digit:
Total sequences = 10 * 10 * 10 * 10 = 10,000
So, there are 10,000 possible sequences for the last four digits.
b. To calculate the probability of getting a 5 or 0 in the last digit, we need to determine the number of favorable outcomes and divide it by the total number of possible outcomes.
The number of favorable outcomes is 2 (either 5 or 0).
The total number of possible outcomes is 10 because there are ten digits (0-9).
Probability = Number of favorable outcomes / Total number of possible outcomes
= 2 / 10
= 1/5
= 0.2
Hence, the probability of getting a 5 or 0 in the last digit is 0.2 or 20%.
c. To calculate the probability of getting two 5's in the last two digits, we need to determine the number of favorable outcomes and divide it by the total number of possible outcomes.
The number of favorable outcomes is 1 (the last two digits being 55).
Again, the total number of possible outcomes is 10 * 10 = 100 (total sequences for the last two digits).
Probability = Number of favorable outcomes / Total number of possible outcomes
= 1 / 100
= 0.01
Hence, the probability of getting two 5's in the last two digits is 0.01 or 1%.