- Each entry in a table of random digits like Table B has probability 0.1 of being a 0, and digits are independent of each other. (5pts) a) Find the probability that a group of five random digits will contain at least one 0. b) Find the probability that a group of five random digits will contain exactly three 0s. c) Suppose we have 40 random digits. What is the mean number of 0s in these 40 digits?
The Correct Answer and Explanation is:
Part a) Find the probability that a group of five random digits will contain at least one 0.
Each digit in the table has a probability of 0.1 of being a 0, and 0.9 of being any other digit (1-9). We can use the complement rule to find the probability of having at least one 0 in five digits. First, we calculate the probability of having no 0s at all, and then subtract that from 1 to find the probability of having at least one 0.
The probability that a single digit is not a 0 is 0.9. Therefore, the probability that none of the five digits are 0s is:P(no 0s)=(0.9)5=0.59049P(\text{no 0s}) = (0.9)^5 = 0.59049P(no 0s)=(0.9)5=0.59049
Thus, the probability of having at least one 0 is:P(at least one 0)=1−P(no 0s)=1−0.59049=0.40951P(\text{at least one 0}) = 1 – P(\text{no 0s}) = 1 – 0.59049 = 0.40951P(at least one 0)=1−P(no 0s)=1−0.59049=0.40951
Part b) Find the probability that a group of five random digits will contain exactly three 0s.
This is a binomial probability problem where the number of trials is 5 (since there are 5 digits), and the probability of success (getting a 0) on each trial is 0.1. The formula for the binomial probability is:P(k successes)=(nk)pk(1−p)n−kP(k \text{ successes}) = \binom{n}{k} p^k (1-p)^{n-k}P(k successes)=(kn)pk(1−p)n−k
where:
- n=5n = 5n=5 is the number of trials,
- k=3k = 3k=3 is the number of successes (0s),
- p=0.1p = 0.1p=0.1 is the probability of success.
The binomial coefficient (nk)\binom{n}{k}(kn) is calculated as:(53)=5!3!(5−3)!=5×42×1=10\binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10(35)=3!(5−3)!5!=2×15×4=10
Now, calculate the probability:P(exactly 3 0s)=(53)(0.1)3(0.9)2=10×(0.1)3×(0.9)2P(\text{exactly 3 0s}) = \binom{5}{3} (0.1)^3 (0.9)^2 = 10 \times (0.1)^3 \times (0.9)^2P(exactly 3 0s)=(35)(0.1)3(0.9)2=10×(0.1)3×(0.9)2P(exactly 3 0s)=10×0.001×0.81=0.0081P(\text{exactly 3 0s}) = 10 \times 0.001 \times 0.81 = 0.0081P(exactly 3 0s)=10×0.001×0.81=0.0081
Part c) Suppose we have 40 random digits. What is the mean number of 0s in these 40 digits?
The number of 0s in 40 digits follows a binomial distribution. The mean of a binomial distribution is given by:μ=n×p\mu = n \times pμ=n×p
where:
- n=40n = 40n=40 is the number of trials,
- p=0.1p = 0.1p=0.1 is the probability of getting a 0.
Thus, the mean number of 0s in 40 digits is:μ=40×0.1=4\mu = 40 \times 0.1 = 4μ=40×0.1=4
So, on average, you can expect 4 zeros in a group of 40 random digits.
Summary of Results:
a) The probability that a group of five random digits will contain at least one 0 is 0.40951.
b) The probability that a group of five random digits will contain exactly three 0s is 0.0081.
c) The mean number of 0s in a group of 40 random digits is 4.
