Question
A box contains five purple pens; three black pens and eight blue pens_ You randomly take two pens out of the box one after another without replacement: Draw a probability tree representing this experimentb. Find the sample space associated with this experimentFind the probability of each outcome of the experimentd. Find the probability that exactly one pen is purpleFind the probability that the second pen is black given that the first one was bluef. Find the probability that the first pen is pur
A box contains five purple pens; three black pens and eight blue pens_ You randomly take two pens out of the box one after another without replacement: Draw a probability tree representing this experiment b. Find the sample space associated with this experiment Find the probability of each outcome of the experiment d. Find the probability that exactly one pen is purple Find the probability that the second pen is black given that the first one was blue f. Find the probability that the first pen is purple given that the second one was purple as well.


Answers
There are six balls in an urn. They are identical except for color. Two are red, three are blue, and one is yellow. You are to draw a ball from the urn, note its color, and set it aside. Then you are to draw another ball from the urn and note its color. (a) Make a tree diagram to show all possible outcomes of the experiment. Label the probability associated with each stage of the experiment on the appropriate branch. (b) Probability Extension Compute the probability for each outcome of the experiment.
Tatiana has X plus two pens in her pocket. So x plus two. Samuel has two X minus one pens in his pocket. We have to find the probability that Tatiana pulls out a blue pen at random, so we know that she has to blue pens as mentioned by part, eh? So the probability is simply gonna be two over two plus X or X plus two part B. Samuel has X minus three blue pens. Probability that he pulls out of blue pen is simply gonna be X minus three over to X minus one part. See, we know we're told that exposed to two X miles when we have to find the probability that they both pull out blue pens at random, all we have to do for this is multiply out the two probabilities of them pulling out blue pens and you get a value of two X minus six over two X squared plus three x plus two part D, the probably that one of them pulls out of blue pen. That's just gonna be the sum of the two minus any overlap. So X minus one over to X minus one, and we're
This problem were given, Ah, three boxes with different colored balls in them. So, uh, is asking us for the probability that the ball on screen So we have to consider all the different ways that we could get green balls. Um, in, uh, so party here. Let's say we got it from box one. Well, that would be a 1/3 chance to get box one. Ah, and there are four out of 20 balls, so four out of 20 are green balls. Alternatively, we can also get it from box too. So that would be another 1/3 chance of getting the box. And then there are eight green balls out of 16. So that would be 8/16. And lastly, uh, 1/3 chance to, um, pick the third bucks. And there are six out of 20 green balls, so I will do some reducing here. That is 1/3 time's 1/5 plus 1/3 times, 1/2 and, ah, that's going to be 1/10 or yeah, let's do 1/3 times 3/10. So 1/15 less. 1/6, plus 1/10 on. That will give us Ah, 1/3 after we, uh, simplify all the fractions and part B. We want the probability of ah having a ball drawn from box to given that it's green. So this is the conditional part. Now the formula would be probability of B two and green over the probability of green. So be too and green is 1/3 and there are eight out of 16 there. So times 1/2 this is the really the second part of ah Partney. And in the bottom we have ah 1/3. That was the answer that we calculated from party and that will give us ah 1/2 as the social.
In Brooklyn. 28. We have an urn that contains 19 bowls. Five of them are red. Six of them are blue on the rest, which is it or green. Now we will select three polls. Well, subjected them randomly for birdie. What is the probability that each of the bulls will be off the same color? After we take the three bulls? We will have, for example, three already or three or Peru or three or green. This means this probability. Verity tohave are all red for red or all blue or all green equals The probability toe have already plus all blue plus or green. This is a repeated experiment. We have three consecutive experiments. The first experiment is to take one example out of 19. Let's calculate the probability toe have all. But for the first experiment, the probability to take 100 is five, divided by 19. And for the second experiment, the probability to get one with will be four divided by 18 because there is one selected for the first experiment. For the third experiment toe have three bowls. The probability to get rid is three, divided by 17. This is for it. The same holds for Be for blue and green. The probability toe have for the first bowl for the first experiment have blue is six, divided by 19. The second will be five, divided by 18. Third will be four, divided by 18, and the same for green is a divided by 19 multiplied boy seven, divided by 18, multiplied by six, divided by 17. This is 17, and by calculating soup, the probability equals probability equals 86. Divided by 969. We can approximate it to be it. Point my in person 48.875 person for Bharti. We want to calculate the probability that each of the bulls will be off different colors, meaning that when we take the first bull, for example, if it's red, then the second will be blue or green, then the third would be green. Then the three colors will be different. There is no torrid. There is no to green or 300 or three blue or any other combinations. We can get the three combinations as blue, red or green or whatever, but the three cars are red, blue and the green. With any order What is this probability? Their probability tohave read, and the blue and the green equals. Let's take the example off having red, blue and green the ability to have read for the first experiment. It's five, but by 19, then the probability toe have blue. In the second experiment will be the number off blue polls. Number of loopholes. It's sex, then it's six but divided by 18. Because there is one selected for the first experience, then the probability have green is a number off green balls, which is it divided by 17. This is for this combination, or for this, or there is another order that is probable toe happen tohave blue Corden trouble then red than green. And this is not the two probabilities there is three factorial or sex probabilities toe order these possibilities. Then we multiply by three factory or six. Then the probability equals 80 divided by three 123 or it equals in percentage 24 points, 77 percent. Now, for the rest off the problem, we will make a little difference. Assembling will be with replacement, which means the first day when we take one symbol or one poll from the first experiment, we were repeated again in tow. The Earth. This will make a little difference for body with the bliss mint. We'll have the same calculations here, but this is not subtracted from one and all these values will be the same value it is. We're not a change 76 seven and six, 18 and 17. Then we will have the probability equals for Bartee five, divided by 19 cube and we have here five Cube plus six a cube divided by 19 cube plus seven cube it Cube. Sorry, It Cube divided by 19 cube and this equals 853 divided by 6859 or in percentage equals 12 point 44%. This is higher than this very which is very loose for both to be the probability, have red and the blue and the green equals. It will be the same here, but this number will be 19 and this number will be 19. Then we have five divided by 19 applied by six divided by 19 developed by a divided by 19 multiplied by the number number of possibilities off ordering these polls, which equals 1440 divided by 6859 and in percentage equals 20 0.99 person. You can see that this value is a little lower than this. Very. It's just logic again, and these are the final answers off our problem.