 Susan Nall

2022-01-06

Two balls are chosen randomly from an urn containing 8 white, 4 black, and 2 orange balls. Suppose that we win 2 for each black ball selected and we lose 2 for each black ball selected and we lose 1 for each white ball selected. Let X denote our winnings. What are the possible values of X, and what are the probabilities associated with each value? John Koga

Random variable X denotes the winning amount. The values of X corresponding to otcomes of the event are: $WW=-2,WO=-1,OO=0,WB=1,BO=2,BB=4$
Total number of choosing 2 balls from the urn 14C2. Number of choosing two white balls is 8C2 as they are only 8 white balls.
$P\left(x=-2\right)=\frac{{n}_{WW}}{{n}_{total}}=\frac{\frac{8}{2}}{\frac{14}{2}}=\frac{28}{91}$
Number of choosing one white and one orange ball is  as both are independent events.
$P\left(X=-1\right)=\frac{{n}_{WO}}{{n}_{total}}=\frac{\frac{8}{1}\frac{2}{1}}{\frac{14}{2}}=\frac{16}{91}$
$P\left(X=1\right)=\frac{{n}_{WB}}{{n}_{total}}=\frac{\frac{8}{1}\frac{4}{1}}{\frac{14}{2}}=\frac{32}{91}$
$P\left(X=0\right)=\frac{{n}_{OO}}{{n}_{total}}=\frac{\frac{2}{2}}{\frac{14}{2}}=\frac{1}{91}$
$P\left(X=2\right)=\frac{{n}_{OB}}{{n}_{total}}=\frac{\frac{2}{1}\frac{4}{1}}{\frac{14}{2}}=\frac{8}{91}$
$P\left(X=4\right)=\frac{{n}_{BB}}{{n}_{total}}=\frac{\frac{4}{2}}{\frac{14}{2}}=\frac{6}{91}$ MoxboasteBots5h Don Sumner

Result:
$\frac{6}{91}$, $\frac{32}{91}$, and $\frac{28}{91}$
Solution:
To solve the problem of selecting balls from an urn and calculating the possible values of winnings (X) along with their associated probabilities, we can follow these steps:
1. Determine the possible outcomes:
- Selecting two black balls
- Selecting one black ball and one white ball
- Selecting two white balls
2. Calculate the winnings for each outcome:
- Selecting two black balls results in a win of 2 for each ball: $2+2=4$.
- Selecting one black ball and one white ball results in a win of 2 for the black ball and a loss of 1 for the white ball: $2-1=1$.
- Selecting two white balls results in a loss of 1 for each ball: $-1-1=-2$.
3. Determine the probabilities associated with each outcome:
- The probability of selecting two black balls can be calculated as the product of the probabilities of selecting the first black ball (4 black balls out of 14 total balls) and the second black ball (3 black balls out of 13 remaining balls): $\left(\frac{4}{14}\right)·\left(\frac{3}{13}\right)=\frac{6}{91}$.
- The probability of selecting one black ball and one white ball can be calculated as the product of the probabilities of selecting a black ball (4 black balls out of 14 total balls) and a white ball (8 white balls out of 13 remaining balls) and multiplying it by 2 (since the order of selection can be either black-white or white-black): $2·\left(\frac{4}{14}\right)·\left(\frac{8}{13}\right)=\frac{32}{91}$.
- The probability of selecting two white balls can be calculated as the product of the probabilities of selecting the first white ball (8 white balls out of 14 total balls) and the second white ball (7 white balls out of 13 remaining balls): $\left(\frac{8}{14}\right)·\left(\frac{7}{13}\right)=\frac{28}{91}$.
4. Summarize the possible values of X and their associated probabilities:
- X can take on the values 4, 1, and -2 with respective probabilities of $\frac{6}{91}$, $\frac{32}{91}$, and $\frac{28}{91}$.
Hence, the possible values of X are 4, 1, and -2, with associated probabilities of $\frac{6}{91}$, $\frac{32}{91}$, and $\frac{28}{91}$ respectively. Eliza Beth13

Let's denote the random variables as follows:
$X=$ the winnings
$W=$ the event of selecting a white ball
$B=$ the event of selecting a black ball
$O=$ the event of selecting an orange ball
We know that there are 8 white balls, 4 black balls, and 2 orange balls in the urn.
Now, we can list all the possible outcomes of selecting two balls:
1. Selecting two white balls:
$P\left(WW\right)=\frac{\left(\genfrac{}{}{0}{}{8}{2}\right)}{\left(\genfrac{}{}{0}{}{14}{2}\right)}=\frac{28}{91}$
In this case, the winnings $X=-2-2=-4$ (since we lose 1 for each white ball selected).
2. Selecting one white ball and one black ball:
$P\left(WB\right)=\frac{\left(\genfrac{}{}{0}{}{8}{1}\right)·\left(\genfrac{}{}{0}{}{4}{1}\right)}{\left(\genfrac{}{}{0}{}{14}{2}\right)}=\frac{32}{91}$
In this case, the winnings $X=-2+2=0$ (since we lose 1 for the white ball and win 2 for the black ball).
3. Selecting one white ball and one orange ball:
$P\left(WO\right)=\frac{\left(\genfrac{}{}{0}{}{8}{1}\right)·\left(\genfrac{}{}{0}{}{2}{1}\right)}{\left(\genfrac{}{}{0}{}{14}{2}\right)}=\frac{16}{91}$
In this case, the winnings $X=-2$ (since we lose 1 for the white ball).
4. Selecting two black balls:
$P\left(BB\right)=\frac{\left(\genfrac{}{}{0}{}{4}{2}\right)}{\left(\genfrac{}{}{0}{}{14}{2}\right)}=\frac{6}{91}$
In this case, the winnings $X=2+2=4$ (since we win 2 for each black ball selected).
5. Selecting one black ball and one orange ball:
$P\left(BO\right)=\frac{\left(\genfrac{}{}{0}{}{4}{1}\right)·\left(\genfrac{}{}{0}{}{2}{1}\right)}{\left(\genfrac{}{}{0}{}{14}{2}\right)}=\frac{8}{91}$
In this case, the winnings $X=2$ (since we win 2 for the black ball).
6. Selecting two orange balls:
$P\left(OO\right)=\frac{\left(\genfrac{}{}{0}{}{2}{2}\right)}{\left(\genfrac{}{}{0}{}{14}{2}\right)}=\frac{1}{91}$
In this case, the winnings $X=0$ (since there are no black or white balls selected).
Now, we can summarize the possible values of $X$ and their associated probabilities:
- $X=-4$ with probability $\frac{28}{91}$
- $X=0$ with probability $\frac{1}{91}+\frac{16}{91}$
- $X=2$ with probability $\frac{8}{91}$
- $X=4$ with probability $\frac{6}{91}$
These are the possible values of $X$ and their respective probabilities. To solve this problem, we can consider all possible combinations of balls that can be chosen from the urn. We will calculate the winnings or losses for each combination and determine the probabilities associated with each outcome.
Let's denote the number of white balls chosen as $W$, the number of black balls chosen as $B$, and the number of orange balls chosen as $O$. We know that $W+B+O=2$, and we need to find the possible values of $X$.
We can start by listing all the possible combinations and calculating the winnings or losses for each combination:
1. If we choose 2 white balls ($W=2$), the winnings will be $X=-2-2=-4$.
2. If we choose 1 white ball and 1 black ball ($W=1$, $B=1$), the winnings will be $X=-1+2=1$.
3. If we choose 1 white ball and 1 orange ball ($W=1$, $O=1$), the winnings will be $X=-1$.
4. If we choose 2 black balls ($B=2$), the winnings will be $X=2+2=4$.
5. If we choose 1 black ball and 1 orange ball ($B=1$, $O=1$), the winnings will be $X=-2$.
6. If we choose 2 orange balls ($O=2$), the winnings will be $X=0$.
Now, let's calculate the probabilities associated with each outcome. Since the balls are chosen randomly, we need to consider the probabilities of each combination occurring.
The total number of ways to choose 2 balls from the urn is given by the combination formula:

1. For the combination with 2 white balls, there are $\left(\genfrac{}{}{0}{}{8}{2}\right)$ ways to choose 2 white balls, and the probability is:
$P\left(W=2\right)=\frac{\left(\genfrac{}{}{0}{}{8}{2}\right)}{\left(\genfrac{}{}{0}{}{14}{2}\right)}$
2. For the combination with 1 white ball and 1 black ball, there are $\left(\genfrac{}{}{0}{}{8}{1}\right)·\left(\genfrac{}{}{0}{}{4}{1}\right)$ ways to choose 1 white ball and 1 black ball, and the probability is:
$P\left(W=1,B=1\right)=\frac{\left(\genfrac{}{}{0}{}{8}{1}\right)·\left(\genfrac{}{}{0}{}{4}{1}\right)}{\left(\genfrac{}{}{0}{}{14}{2}\right)}$
3. For the combination with 1 white ball and 1 orange ball, there are $\left(\genfrac{}{}{0}{}{8}{1}\right)·\left(\genfrac{}{}{0}{}{2}{1}\right)$ ways to choose 1 white ball and 1 orange ball, and the probability is:
$P\left(W=1,O=1\right)=\frac{\left(\genfrac{}{}{0}{}{8}{1}\right)·\left(\genfrac{}{}{0}{}{2}{1}\right)}{\left(\genfrac{}{}{0}{}{14}{2}\right)}$
4. For the combination with 2 black balls, there are $\left(\genfrac{}{}{0}{}{4}{2}\right)$ ways to choose 2 black balls, and the probability is:
$P\left(B=2\right)=\frac{\left(\genfrac{}{}{0}{}{4}{2}\right)}{\left(\genfrac{}{}{0}{}{14}{2}\right)}$
5. For the combination with 1 black ball and 1 orange ball, there are $\left(\genfrac{}{}{0}{}{4}{1}\right)·\left(\genfrac{}{}{0}{}{2}{1}\right)$ ways to choose 1 black ball and 1 orange ball, and the probability is:
$P\left(B=1,O=1\right)=\frac{\left(\genfrac{}{}{0}{}{4}{1}\right)·\left(\genfrac{}{}{0}{}{2}{1}\right)}{\left(\genfrac{}{}{0}{}{14}{2}\right)}$
6. For the combination with 2 orange balls, there is only 1 way to choose 2 orange balls, and the probability is:
$P\left(O=2\right)=\frac{1}{\left(\genfrac{}{}{0}{}{14}{2}\right)}$
Now we have calculated the probabilities associated with each outcome. To summarize, the possible values of $X$ and their corresponding probabilities are:
- $X=-4$ with probability $P\left(W=2\right)=\frac{\left(\genfrac{}{}{0}{}{8}{2}\right)}{\left(\genfrac{}{}{0}{}{14}{2}\right)}$
- $X=1$ with probability $P\left(W=1,B=1\right)=\frac{\left(\genfrac{}{}{0}{}{8}{1}\right)·\left(\genfrac{}{}{0}{}{4}{1}\right)}{\left(\genfrac{}{}{0}{}{14}{2}\right)}$
- $X=-1$ with probability $P\left(W=1,O=1\right)=\frac{\left(\genfrac{}{}{0}{}{8}{1}\right)·\left(\genfrac{}{}{0}{}{2}{1}\right)}{\left(\genfrac{}{}{0}{}{14}{2}\right)}$
- $X=4$ with probability $P\left(B=2\right)=\frac{\left(\genfrac{}{}{0}{}{4}{2}\right)}{\left(\genfrac{}{}{0}{}{14}{2}\right)}$
- $X=-2$ with probability $P\left(B=1,O=1\right)=\frac{\left(\genfrac{}{}{0}{}{4}{1}\right)·\left(\genfrac{}{}{0}{}{2}{1}\right)}{\left(\genfrac{}{}{0}{}{14}{2}\right)}$
- $X=0$ with probability $P\left(O=2\right)=\frac{1}{\left(\genfrac{}{}{0}{}{14}{2}\right)}$

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