How to calculate the probability of specific events in probability questions?

  • Thread starter whitehorsey
  • Start date
  • Tags
    Probability
To get a group of 3 we divide by 3*2*1 because we can pick the same group of 3 people in 3*2*1 ways. So the answer is 15*14*13/3*2*1 = 1820In summary, there are 1820 ways to form a committee of 3 or 4 people from a staff of 15. To find the probability that Henri, Marcy, and Jorge are all on the committee, we take the number of possible committees with these three people (13) and divide it by the total number of possible committees (1820), giving us a probability of 13/1820.For the second question, the probability of getting
  • #1
whitehorsey
192
0
1. In how many ways could a committee of 3 or 4 people be formed from a staff of 15.
I got the answer for this which is 1820 ways, but then it asks another question later on.
For the situation of Exercise 17, what is the probability that Henri, Marcy, and Jorge are all on the committee?



2. n!/r!(n-r)!


3. 3!/1820 + 2!/1820 + 1!/1820
= 9/1820
but my answer is wrong it should be 13/1820. How do you do this problem?


2nd question

1. Two cards are drawn in succession from an ordinary deck. The first is not replaced before the second drawn. Find the probability of each event.
Both are red hearts


2. no needed formulas


3. 26/52 * 25/51
= 650/7652
= 325/3826 but my answer is wrong it should be 25/102. How do you do this problem? Thank You!
 
Physics news on Phys.org
  • #2
For the first one, I don't really understand what you are doing with the 3! + 2! + 1!.
I would simply count how many out of the 1820 possible committees contain all those three persons. Note that in those 1820 you have already discarded the order in which they are elected. So of all the possibilities where the committee is formed of 3 people, there is just 1 which contains those three specific people. How many committees of 4 people are there which contain those three?

For the second one, you have made a calculation error. The answer is indeed 26/52 * 25/51, but 52 * 51 is 2652, not 7652.
 
  • #3
CompuChip said:
For the first one, I don't really understand what you are doing with the 3! + 2! + 1!.
I would simply count how many out of the 1820 possible committees contain all those three persons. Note that in those 1820 you have already discarded the order in which they are elected. So of all the possibilities where the committee is formed of 3 people, there is just 1 which contains those three specific people. How many committees of 4 people are there which contain those three?

For the second one, you have made a calculation error. The answer is indeed 26/52 * 25/51, but 52 * 51 is 2652, not 7652.

hmmm I don't really get what your saying on the first explanation like if there is only 1 possibility that contains those 3 people then the answer is 1/1820?

For the second one which is 650/2652 I can't seem to simplify it into getting 25/102. Is it because my answer is wrong or I'm not simplifying right.
I first divide by 2 getting 325/1326. ---- This seems to be all I can do.​
 
  • #4
No, I said that there is one possibility in the set of committees with just three people and you need to take into account the committees with 4 people as well.

For the second one: try a factor 13.
 
  • #5
For the second question, 26/52 is the probability of a red card, not of a "red heart".
All hearts are red, so not sure if they mean "what is the probability of getting two hearts successively". If it does, your numbers are wrong.
 
  • #6
CompuChip said:
No, I said that there is one possibility in the set of committees with just three people and you need to take into account the committees with 4 people as well.

For the second one: try a factor 13.

o so is it because out of the 15 committee memebers they are 3 that are in a group of 4
so i multiply 4 *3 = 12 and there is one group of 3
= 12/1820 + 1/1820
= 13/1820 Is my reasoning right?

I get the second one now.
Thank you for helping me! I have a test coming up and I want to do good on it. :)
 
  • #7
I'm really bad at these questions but this is my current understanding of the problem.

If you do not repeat events (which is the case in the 1st question) then you will be looking at getting 15*14*13 places because we can choose the first person in 15 ways the second in 14 ways the third in 13 ways.
 

1. What is probability and how is it calculated?

Probability is the measure of how likely an event is to occur. It is calculated by dividing the number of desired outcomes by the total number of possible outcomes.

2. What are the different types of probability?

There are three main types of probability: theoretical, experimental, and subjective. Theoretical probability is based on mathematical principles and assumes that all outcomes are equally likely. Experimental probability is based on data collected from experiments or observations. Subjective probability is based on personal opinions or beliefs.

3. How is probability used in real life?

Probability is used in many real-life situations, such as predicting the weather, determining the likelihood of winning a game or lottery, and making financial decisions. It is also used in fields like insurance, medicine, and engineering to assess risk and make informed decisions.

4. What is the difference between independent and dependent events?

Independent events are events where the outcome of one event does not affect the outcome of the other event. Dependent events are events where the outcome of one event does depend on the outcome of the other event. For example, flipping a coin twice is an independent event, while drawing two cards from a deck without replacement is a dependent event.

5. How can I improve my understanding of probability?

To improve your understanding of probability, it is helpful to practice solving problems and working with different types of probability. It is also important to understand basic mathematical concepts and terminology, such as combinations and permutations. Additionally, seeking out resources such as textbooks, online tutorials, and practice exercises can also be beneficial.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
12
Views
1K
  • Precalculus Mathematics Homework Help
Replies
22
Views
1K
  • Precalculus Mathematics Homework Help
Replies
6
Views
868
  • Precalculus Mathematics Homework Help
Replies
29
Views
2K
  • Precalculus Mathematics Homework Help
Replies
12
Views
1K
  • Precalculus Mathematics Homework Help
Replies
7
Views
4K
  • Precalculus Mathematics Homework Help
Replies
12
Views
2K
  • Precalculus Mathematics Homework Help
Replies
6
Views
1K
  • Precalculus Mathematics Homework Help
Replies
7
Views
1K
  • Precalculus Mathematics Homework Help
Replies
6
Views
2K
Back
Top