# A bag contains 2 red checkers and 6 black checkers. A checker is selected, kept out of the bag, and then another checker is selected. What is P(black, then red)?

Question
A bag contains 2 red checkers and 6 black checkers. A checker is selected, kept out of the bag, and then another checker is selected. What is P(black, then red)?

2020-12-23
There are 6 black checkers out of the 8 checkers so the probability of selecting a black checker on the first draw is:
P(black first)=$$\displaystyle\frac{{6}}{{8}}=\frac{{3}}{{4}}$$
Since the checker is not replaced, then there are now 7 checkers left so the probability of selecting a red checker on the second draw is:
P(red second)=$$\displaystyle\frac{{2}}{{7}}$$
So, the probability of drawing a black checker first, then a red checker second is:
P(black, the red)=$$\displaystyle{\left(\frac{{3}}{{4}}\right)}\cdot{\left(\frac{{2}}{{7}}\right)}$$
P(black, the red)=$$\displaystyle\frac{{6}}{{28}}$$
P(black, the red)=$$\displaystyle\frac{{3}}{{14}}$$

### Relevant Questions

A gambling book recommends the following "winning strategy" for the game of roulette. It recommends that a gambler bet $1 onred. If red appears (which has probablity 18/38), then the gamblershould take her$1 profit and quit. If the gambler loses this bet (which has probablity 20/38 of occurring), she should make additional \$1 bets on red on each of the next two spins of the roulette wheel and then quite. Let X denote the gambler's winnings when she quites.
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The table below shows the number of people for three different race groups who were shot by police that were either armed or unarmed. These values are very close to the exact numbers. They have been changed slightly for each student to get a unique problem.
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White - 1176
Hispanic - 378
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White - 67
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White - 1243
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c) In order for two variables to be Independent of each other, the P $$(A and B) = P(A) \cdot P(B) P(A and B) = P(A) \cdot P(B).$$
This just means that the percentage of times that both things happen equals the individual percentages multiplied together (Only if they are Independent of each other).
Therefore, if a person's race is independent of whether they were killed being unarmed then the percentage of black people that are killed while being unarmed should equal the percentage of blacks times the percentage of Unarmed. Let's check this. Multiply your answer to part a (percentage of blacks) by your answer to part b (percentage of unarmed).
Remember, the previous answer is only correct if the variables are Independent.
d) Now let's get the real percent that are Black and Unarmed by using the table?
If answer c is "significantly different" than answer d, then that means that there could be a different percentage of unarmed people being shot based on race. We will check this out later in the course.
Let's compare the percentage of unarmed shot for each race.
e) What percent are White and Unarmed?
f) What percent are Hispanic and Unarmed?
If you compare answers d, e and f it shows the highest percentage of unarmed people being shot is most likely white.
Why is that?
This is because there are more white people in the United States than any other race and therefore there are likely to be more white people in the table. Since there are more white people in the table, there most likely would be more white and unarmed people shot by police than any other race. This pulls the percentage of white and unarmed up. In addition, there most likely would be more white and armed shot by police. All the percentages for white people would be higher, because there are more white people. For example, the table contains very few Hispanic people, and the percentage of people in the table that were Hispanic and unarmed is the lowest percentage.
Think of it this way. If you went to a college that was 90% female and 10% male, then females would most likely have the highest percentage of A grades. They would also most likely have the highest percentage of B, C, D and F grades
The correct way to compare is "conditional probability". Conditional probability is getting the probability of something happening, given we are dealing with just the people in a particular group.
g) What percent of blacks shot and killed by police were unarmed?
h) What percent of whites shot and killed by police were unarmed?
i) What percent of Hispanics shot and killed by police were unarmed?
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j) Why do you believe this is happening?
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The problem reads: Suppose $$\displaystyle{P}{\left({X}_{{1}}\right)}={.75}$$ and $$\displaystyle{P}{\left({Y}_{{2}}{\mid}{X}_{{1}}\right)}={.40}$$. What is the joint probability of $$\displaystyle{X}_{{1}}$$ and $$\displaystyle{Y}_{{2}}$$?
This is how I answered it. P($$\displaystyle{X}_{{1}}$$ and $$\displaystyle{Y}_{{2}}$$) $$\displaystyle={P}{\left({X}_{{1}}\right)}\times{P}{\left({Y}_{{1}}{\mid}{X}_{{1}}\right)}={.75}\times{.40}={0.3}.$$
What I don't understand is how do you get the $$\displaystyle{P}{\left({Y}_{{1}}{\mid}{X}_{{1}}\right)}$$? I am totally new to Statistices and I need to understand each part of the process in order to get the whole concept. Can anyone help me to understand why the P and X exist and what they represent?