brook633
20.12.2019 •
Mathematics
Suppose that 3 balls are chosen without replacement from an urn consisting of 3 white and 7 red balls. assume moreover that the white balls are labeled 1; 2; 3. let xi = 1 if the i-th white ball is chosen among the 3 selected balls, and 0 otherwise. find the pmf of (x1; x2).
Solved
Show answers
More tips
- G Goods and services How to sew a ribbon: Tips for beginners...
- F Food and Cooking How to Make Mayonnaise at Home? Secrets of Homemade Mayonnaise...
- C Computers and Internet Which Phone is Best for Internet Surfing?...
- F Food and Cooking Everything You Need to Know About Pasta...
- C Computers and Internet How to Choose a Monitor?...
- H Horoscopes, Magic, Divination Where Did Tarot Cards Come From?...
- S Style and Beauty How to Make Your Lips Fuller? Ideas and Tips for Beautiful Lips...
- C Computers and Internet How to Learn to Type Fast?...
Answers on questions: Mathematics
- M Mathematics Beverly is 10 years old and he has collected 15 sea shells. Sharon is 8 years old and has collected 18 sea shells. Brad is Beverly s twin brother. Brad collected 32...
- M Mathematics How do you use 4 8 2 7 4 9...
- M Mathematics What is the answer to 3x=5(m+3)-3...
- M Mathematics Giving please help Do not round your answer. What is 8% of 200?...
- M Mathematics What is Y=-2x+50 graphed...
- M Mathematics What is the image point of (3,-4) after a translation right 1 unit and up 4 units?...
- M Mathematics How do you use 4 8 2 7 4 9 to get 100?...
- M Mathematics The table below shows the median selling price of houses in the 1990s. Draw a scatter plot on the data...
- M Mathematics Who would date me ?????...
- M Mathematics Hi welcome to Chick-Fil-A what can i get for u today??? free points...
Ответ:
P(X1=1, X2=1) = 1/15
P(X1=1, X2=0) = 7/30
P(X1=0, X2=1) = 7/30
P(X1=0, X2=0) = 7/15
Step-by-step explanation:
Let Xi = 1 if the i-th white ball is selected. In this question the 3 white balls are marked 1,2 and 3.
We need to know the possible combination between X1 and X2 i.e. for the white ball 1 and 2 being chosen in the event.
We also need to note that the event is dependent which that after a ball is being chosen, it will not be put back hence affecting the probability of picking the next ball.
Consider all the possible combination between X1 and X2
a) both being chosen P(X1=1, X2=1)
= (3/10) x (2/9) = 1/15
Note that the first probability is the probability before any ball is being picked. The chances for ball white to be pick is 3/10 (3 white ball from the total 10 balls).
After 1 white ball being selected, that ball is not again out back into the urn making white ball 2 and total ball 9. Hence probability of picking another white ball is 2/9
b) only X1 chosen P(X1=1, X2=0)
= (3/10) x (7/9) = 7/30
After the white ball was picked, the probability of white not being pick again is the same as red being picked. Since there is still 7red balls and a total of 9 balls, the probability is 7/9
c) only X2 chosen P(X1=0, X2=1)
= (7/10) x (3/9) = 7/30
The white is not being picked first, making the probability of picking red is 7/10. Then the probability of white being picked is 3/9
d) both not chosen
P(X1=0, X2=0)
= (7/10) x (6/9) = 7/15
In other word only red being chosen. So the first probability is 7 red out of 10 balls (7/10), and the next red ball being picked next is 6/9
Ответ:
2 times pi (3.14) is about 6.28
radius half of the diameter so half of 10 is 5
so 5 times 6.28 is
31.4.