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Two confusing questions on Probability

  • Thread starter zorro
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  • #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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Answers and Replies

  • #2
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Anyone??
Please help me.
 
  • #3
verty
Homework Helper
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Try to use combinations instead. In the first question, how many combinations are there for choosing 4 people but no couple?
 
  • #4
LeonhardEuler
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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
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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:

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

This is what I can think of other than my earlier attempt.
Keep thinking like that.
 
  • #7
LeonhardEuler
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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
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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
LeonhardEuler
Gold Member
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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
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So the answer 15/39 is incorrect?
 
  • #11
LeonhardEuler
Gold Member
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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
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Aargh!! 15/39 = 5/13 :redface:
Thank you very much :smile:
 

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