I need help with this question.?

Question

5 people were wearing visors at a game. When a sudden wind blew their visors on to the field. A boy picked them up [without knowing which visor belonged to which person] and gave each person a visor. What chance [percentage] would there be for the boy to have exactly four visors correctly placed?

Answer 1

The chance that exactly *four* visors are placed correctly is 0%.

Tell me how you are going to place *four* correctly and not have the *fifth* placed correctly too?

Remember the question asks the probability that *exactly* four are correct, so there is no way to have exactly four correct.

This is a trick question and the answer is 0%.

Answer 2

Your question specifies exactly four correctly in place; of course, if you have four in place, you must have the fifth one in place as well. There is no choice on the last one.

You have five people and five visors. The only correct outcome, the success outcome, is the one where visor 1 fits person 1, visor 2 fits person 2, etc.

Further, there are 25 possible outcomes before the first attempt to match is made. This is demonstrated by a m X m, 5 X 5 table, where the rows, say, are the persons, and the columns are the visors. Each of the 25 row-column elements in the table has an equal probability of happening at random.

The success outcome lies only on the main diagonal of the 5 X 5 table, where person 1 has visor 1, etc. Anywhere else off the main diagonal represents a failure to get all five person-visor matches matched successfully. Thus, since there are five main diagonal elements representing a success n(s) = 5, the number of elements in the table representing a match of the right visor with the right person.

The probability of making a match on the first attempt is P(s) = n(s)/N = 5/25. That is, there are five possible matches along the diagonal of a m X m = 5 X 5 = 25 element table of possible outcomes.

If that first attempt was a success, then the probability of making a match on the second attempt is P(s|1) = n(s|1)/(m-1 X m-1) = 4/(4 X 4) = 4/16; where P(s|1) is the probability of a match given the first (1) attempt was a success.

n(s|1) is the number of possible successes (matches) remaining after the first match was made. Since one match was already made, there are only four more possible matches to make, which is why n(s|1) = 4 for the second attempt.

Also, since the one person and one visor are no longer a possibility, but a fact, the m X m table is now reduced to m-1 X m-1 = 4 X 4 = 16 possibilities with the remaining visors and people. That is to say, after a successful attempt, the row and column associated with that success on the diagonal are no longer elligible for more attempts.

Following this line of thought, the probability of making five matches out of five attempts is P(s AND s AND s AND s AND s) = P(s|0)P(s|1)P(s|2)P(s|3)P(s|4) = (5/25)(4/16)(3/9)(2/4)(1); where the (P(s|4)=1) is a fact, not a probability, that you will match that last visor because once the first four are matched the fifth one has to be matched.

I presume you've noticed that each individual attempt probability is just P(s|l) = 1/(m-l); where m-l is the number of rows or columns left in the table for the next attempt. For example, after two successes l = 2; so that m - l = 5 - 2 = 3. In which case, P(s|2) = 1/3 and P(s|3) = 1/(5 - 3) = 1/2 and P(s|4) = 1/1. I mention this because now you can breeze through this kind of problem by inspection.

If you need a single probability number, just multiply the individual attempt probabilities I gave you above.

Answer 3

1 to 5

Answer 4

For one person, the probability that the visor is correct is 1/5 or 20%.

For four people, the probability that the visors are correct is (1/5)^4 or 0.16%.

Answer 5

Here is my guess.

there are 5x4x3x2x1 possibilities for returning the visors (or, 120)

There are 5 ways that four people could have the correct visor,

So my GUESS (AND i EMPHASIZE GUESS, SINCE PROBABILITIES ISN'T MY BEST SUBJECT)

my guess is 5/120

Now convert this to a percentage by dividing 5 by 120, then multiplying by 100.

Answer 6

do 1/5*1/5*1/5*1/5 note this is the same odds as that all 5 are in the right spot, seeing as that the if 4 are correct, and there are 5 people, then the last one couldnt be wrong. if you have a graphing calculator use the combination and permutation functions found under the math button. itl save you so much time on that unit.

Answer 7

there are 120 ways he could give back the hats
5X4X3X2X1 or 5! 5 factorial the chance of him giving all the hats back to the right owners 1 out of 120 or .833%

Answer 8

I guess 1 in 5 since
there are 5 persons and each person only has one visor.

but dont write that down until you get info from other persons
I dont love math....

Answer 9

he has a 25%

Answer 10

20%