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31 Jul 2013, 04:51
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Another way:
4/4 * 3/4 * 2/4 * 1/4 = 3/32 = 0.09375 => 9% approx
4/4 * 3/4 * 2/4 * 1/4 = 3/32 = 0.09375 => 9% approx
27 Jan 2014, 04:52
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tarek99
Although this is a very basic question and there are several responses i would like to reply elaborately.
There are 4 persons and each has to choose different numbers. we can have 2 approach.
1. Plain and simple counting
Lets say there are 4 persons : A,B,C,D and they need to choose 4 numbers such that no 2 person choose the same number.
How many choice does A have, if he is the first one to choose. It should be 4. There are 4 numbers and A does not have any constraint. So A can choose in 4 ways
How many choice does B have, if he is the second one to choose. It should be 3. There are in all 4 numbers. But A has already chosen 1, so B has only 3 numbers to choose from.
How many choice does C have, if he is the Third one to choose. It should be 2. There are in all 4 numbers. But A and B has already chosen 2 numbers, so C has only 2 numbers to choose from.
How many choice does D have, if he is the fourth one to choose. It should be 1. There are in all 4 numbers. But A, B and C has already chosen 3 numbers, so D has only 1 number left to choose from.
Now,
Probability of A = Number of favorable outcomes / Total number of outcomes.
Number of favorable outcome = 4 --> As we calculated above.
Total number of outcomes = 4. There are only 4 numbers to choose from.
so P(A) = 4/4
Similarly P(B) = 3/4, P(C) = 2/4, P(D) = 1/4
Now all the persons have to choose the ball for the same event. Its like A chooses AND B chooses AND C chooses AND D chooses. Since we use 'and' we need to multiply.
Hence probability = 4/4 * 3/4*2/4*1/4 = 3/32 ~ 9%
2. Applying combinations
Lets say there are 4 persons : A,B,C,D and they need to choose 4 numbers such that no 2 person choose the same number.
Now this is an unordered arrangement. It does not matter if the 4 chosen numbers are 1,2,3,4 or 2,1,3,4. We are only concerned with the fact that numbers chosen should be different. We are not bothered about a particular order and hence combination applies.
Having said that,
In how many ways can person A choose 1 number from available 4 number? Try it.
4c1
3C1
C : 2C1 and D 1C1
what will be P(A)?
4c1/4c1. Total number of outcomes = 4C1
B : 3C1/4c1, C : 2C1/4c1, D : 1C1/4c1
P(A) * P(B) * P (C) * P(D) = 4C1/4C1 * 3C1/4C1 * 2C1/4C1 * 1C1/4C1 = 3/32
30 Mar 2016, 12:04
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Primarily no of ways we can select a number for a person = 4 ways since there are 4 nos
so if different nos are to be chosen
if 1st person selects a no then he has all 4 chances
next how many are available fr 2nd one........obviously 3 remaining
so second one has 3 ways
similarly third one has 2 ways
final person is left with only one number
since all of this happen correspondingly and are inter related events they are multiplied
and hence numerator of favorable outcomes is 4*3*2*1 = 24
now we have to consider all the possible cases where all nos can be same or different
that means irrespective of what one chooses does not bother other one and everyone can choose anyone of the 4 nos
hence 4 is multiplied for 4 times for all possible cases
hence
\(prob=\frac{fav}{total}=\frac{{24}}{{4*4*4*4}}\)
since all the choices are in percentages, we convert his one also in percentage by multiplying the prob by 100 i.e., 9% approximately.
hence A is the answer
so if different nos are to be chosen
if 1st person selects a no then he has all 4 chances
next how many are available fr 2nd one........obviously 3 remaining
so second one has 3 ways
similarly third one has 2 ways
final person is left with only one number
since all of this happen correspondingly and are inter related events they are multiplied
and hence numerator of favorable outcomes is 4*3*2*1 = 24
now we have to consider all the possible cases where all nos can be same or different
that means irrespective of what one chooses does not bother other one and everyone can choose anyone of the 4 nos
hence 4 is multiplied for 4 times for all possible cases
hence
\(prob=\frac{fav}{total}=\frac{{24}}{{4*4*4*4}}\)
since all the choices are in percentages, we convert his one also in percentage by multiplying the prob by 100 i.e., 9% approximately.
hence A is the answer
