Question

assignment 1 replacement: problem 8 (1 point) a group of 15 numbers of which 6 are even and 9 are odd. four numbers are randomly selected and not replaced. the probability that there will be 3 odd numbers among them is a. (\\(\frac{9}{15}\\))³\\(\frac{6}{15}\\) b. \\(\frac{\binom{9}{3}\binom{6}{1}}{\binom{15}{4}}\\) c. \\(\frac{\binom{9}{3}\binom{6}{1}}{\binom{15}{4}}\\) d. \\(\frac{\binom{9}{3}\binom{6}{1}}{\binom{15}{4}}\\) four numbers are randomly selected one at a time, recorded and replaced before the next is chosen. the probability that there will be 3 odd numbers among them is a. (\\(\frac{9}{15}\\))³\\(\frac{6}{15}\\) b. \\(\frac{\binom{4}{3}(\frac{9}{15})^3(\frac{6}{15})^1}{\binom{15}{4}}\\) c. 4(\\(\frac{9}{15}\\))\\(\frac{6}{15}\\) d. (\\(\frac{9}{15}\\))³\\(\frac{6}{15}\\)

Explanation:

Step1: Calculate non - replacement case

We use the combination formula $C(n,k)=\frac{n!}{k!(n - k)!}$. The total number of ways to choose 4 numbers out of 15 is $C(15,4)=\frac{15!}{4!(15 - 4)!}=\frac{15!}{4!11!}$. The number of ways to choose 3 odd numbers out of 9 is $C(9,3)=\frac{9!}{3!(9 - 3)!}=\frac{9!}{3!6!}$, and the number of ways to choose 1 even number out of 6 is $C(6,1)=\frac{6!}{1!(6 - 1)!}=\frac{6!}{1!5!}$. The probability of choosing 3 odd and 1 even number (non - replacement) is $\frac{C(9,3)\times C(6,1)}{C(15,4)}$, so the answer for the non - replacement part is D.

Step2: Calculate replacement case

This is a binomial probability problem. The binomial probability formula is $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $n$ is the number of trials, $k$ is the number of successes, and $p$ is the probability of success on a single trial. Here, $n = 4$, $k=3$, the probability of choosing an odd number $p=\frac{9}{15}$, and the probability of choosing an even number $1 - p=\frac{6}{15}$. $C(4,3)=\frac{4!}{3!(4 - 3)!}=4$. So $P(X = 3)=C(4,3)\times(\frac{9}{15})^{3}\times\frac{6}{15}=4\times(\frac{9}{15})^{3}\times\frac{6}{15}$, and the answer for the replacement part is C.

Answer:

1. D. $\frac{C(9,3)\times C(6,1)}{C(15,4)}$
2. C. $4(\frac{9}{15})^{3}\frac{6}{15}$