Each entry in a table of random digits like Table D has probability 0.1 of being a
and the digits are independent of one another. If many lines of 40 random digits are selected, the mean and standard deviation of the number of 0 s will be approximately (a) mean
standard deviation
. (b) mean
standard deviation
(c) mean
standard deviation
. (d) mean
standard deviation
. (e) mean
standard deviation
.
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.
