Two confusing questions on Probability

In summary, the probability of at least one couple being selected in a lucky draw with 8 couples and 4 prizes is 5/13. The probability of at least one pair being selected among 8 pairs of shoes when 8 shoes are randomly chosen is 344/351. The method used to calculate this probability is by finding the probability of not selecting any couples or selecting exactly 4 pairs and subtracting it from 1. This involves using combinations to find the number of ways to select 4 people without choosing any couples and then multiplying it by 2 for each couple. The final probability is then calculated by dividing this number by the total number of ways to choose 4 people out of 16.
  • #1
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Homework Statement



I am confused in the approach to two similar questions. So I am posting both-

1) A team of 8 couples (husband and wife) attend a lucky draw in which 4 persons are picked up for a prize. The probability that there is atleast one couple is?

2) 8 pairs of shoes are in a closet. 8 shoes are selected at random. The probability that there will be at least one pair and atmost 3 pairs among selected shoes is?


The Attempt at a Solution



1) Required probability = 1 - probability of not selecting any couple for prize
= 1 - (16 x 14 x 12 x 10)/(16P4)
= 5/13

2) Required probability = 1 - probability of selecting no pair - probability of selecting 4 pairs
= 1 - (16 x 14 x 12 x 10 ...x 2)/(16P8) - (16 x 1 x 14 x 1...10 x 1)/(16P8)
= 344/351

Is the method right?
1)15/39 2)1 - 326/16C8
 
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  • #2
Anyone??
Please help me.
 
  • #3
Try to use combinations instead. In the first question, how many combinations are there for choosing 4 people but no couple?
 
  • #4
Abdul Quadeer said:
2) Required probability = 1 - probability of selecting no pair - probability of selecting 4 pairs
= 1 - (16 x 14 x 12 x 10 ...x 2)/(16P8) - (16 x 1 x 14 x 1...10 x 1)/(16P8)
= 344/351

Is the method right?
1)15/39 2)1 - 326/16C8
This part is not right. Finding between 1 and 3 pairs is not the same as 1 - P(selecting no pairs) - P(selecting exactly 4 pairs). Here you have excluded only the cases where no pairs are found and where exactly 4 pairs are found. So this would be P(not selecting either 0 or 4 pairs) if you had calculated it correctly. Also, P(selecting exactly 4 pairs) is not
(16 x 1 x 14 x 1...10 x 1)/(16P8)
because you do not need to pick each pair with both shoes in a row.
 
  • #5
vertigo said:
Try to use combinations instead. In the first question, how many combinations are there for choosing 4 people but no couple?

16 x 14 x 12 x 10/(16C4)

First we select one out of 16 persons in 16C1 ways. Next person can be selected in 14C1 ways as the 15th one might be the husband/wife of the first person selected. Rest follows.
But this does not give a valid result so I used permutations :biggrin:

LeonhardEuler said:
This part is not right. Finding between 1 and 3 pairs is not the same as 1 - P(selecting no pairs) - P(selecting exactly 4 pairs). Here you have excluded only the cases where no pairs are found and where exactly 4 pairs are found. So this would be P(not selecting either 0 or 4 pairs) if you had calculated it correctly. Also, P(selecting exactly 4 pairs) is not
(16 x 1 x 14 x 1...10 x 1)/(16P8)
because you do not need to pick each pair with both shoes in a row.

(Either 0 or 4 pairs) and (0 and 4 pairs) are both same as they are mutually exclusive

P(selecting exactly 4 pairs) = 8C4/16C8

This is what I can think of other than my earlier attempt.
 
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  • #6
Abdul Quadeer said:
P(selecting exactly 4 pairs) = 8C4/16C8

This is what I can think of other than my earlier attempt.

Keep thinking like that.
 
  • #7
Abdul Quadeer said:
P(selecting exactly 4 pairs) = 8C4/16C8

This is what I can think of other than my earlier attempt.

This equation is correct. Now that I think about it, all of your descriptions of the required probability are correct. I was thinking that saying "not 0 or 4" was not enough because there are 8 pairs, but you're right because you can only select up to 4 pairs if you choose 8 shoes. The only problem was with your description of the probability of selecting 4 pairs, which you have now corrected, so everything should be fine.
 
  • #8
LeonhardEuler said:
This equation is correct. Now that I think about it, all of your descriptions of the required probability are correct. I was thinking that saying "not 0 or 4" was not enough because there are 8 pairs, but you're right because you can only select up to 4 pairs if you choose 8 shoes. The only problem was with your description of the probability of selecting 4 pairs, which you have now corrected, so everything should be fine.

Do you mean that this is correct-
P(no pair) = (16 x 14 x 12 x 10 ...x 2)/(16P8)

If yes then I feel it really weird to use permutations here and combinations in selecting exactly 4 pairs. There must be some other way of using combinations in it.
 
  • #9
Yes, they are both right. It's fine and simpler to use permutations in the first case. Basically, you want to know the fraction of outcomes with no pairs. You can consider outcomes to be either ordered quintuples of names of the people, or unordered ones. The fraction will be the same. If you wanted to take the unordered point of view, here is how I would do it:

To select no couples among 4 choices, you first need to pick 4 couples among the 8. The number of ways to do this is
[tex]^{8}C_{4}[/tex]
Next, you need to pick a person from each couple to be selected. There are 2 ways of doing this per couple, so you pick up a factor of 24. The total number of ways of picking 4 people out of 16 is
[tex]^{16}C_{4}[/tex]
Therefore the probability of picking no couples is:
[tex]\frac{^{8}C_{4}*2^{4}}{^{16}C_{4}} = \frac{8}{13}[/tex]
 
  • #10
So the answer 15/39 is incorrect?
 
  • #11
Abdul Quadeer said:
So the answer 15/39 is incorrect?

Why would you say that? The calculation I just showed was that the probability of no couples was
[tex]\frac{8}{13}[/tex]
so the probability of at least 1 couple would be
[tex]1-\frac{8}{13} = \frac{5}{13} = \frac{15}{39}[/tex]
 
  • #12
Aargh! 15/39 = 5/13 :redface:
Thank you very much :smile:
 

1. What is the difference between independent and dependent events in probability?

Independent events are events where the outcome of one event does not affect the outcome of the other event. This means that the probability of the second event occurring remains the same, regardless of whether or not the first event occurred. Dependent events, on the other hand, are events where the outcome of one event does affect the outcome of the other event. This means that the probability of the second event occurring changes based on whether or not the first event occurred.

2. How do you calculate the probability of an event?

The probability of an event is calculated by dividing the number of favorable outcomes by the total number of possible outcomes. This can be represented by the formula P(event) = number of favorable outcomes / total number of possible outcomes. Probability is usually expressed as a decimal, percentage, or fraction.

3. What is the difference between theoretical probability and experimental probability?

Theoretical probability is the probability of an event occurring based on mathematical calculations and assumptions. It is based on the assumption that all outcomes are equally likely. Experimental probability, on the other hand, is the probability of an event occurring based on actual experiments or data collected from real-life situations. It may differ from theoretical probability due to factors such as chance or human error.

4. What is the difference between permutation and combination in probability?

Permutation is an arrangement of objects where order matters. This means that the arrangement of the objects affects the outcome. Combination, on the other hand, is a selection of objects where order does not matter. This means that the selection of objects does not affect the outcome. In probability, permutation is used to calculate the number of ways to arrange a set of objects, while combination is used to calculate the number of ways to select a subset of objects from a larger set.

5. How is probability used in real-life situations?

Probability is used in real-life situations to make predictions and decisions based on the likelihood of certain events occurring. It is used in fields such as statistics, finance, and science to analyze and interpret data. In everyday life, probability is used to make decisions such as buying a lottery ticket, predicting the weather, or choosing the best route to avoid traffic. It is also used in risk assessment and decision-making in various industries.

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