Question native Fatima, a parent:
Six world call castle A,B,C,D,E,F are randomly split into three groups of two,find the probability the the below event(do not impose undesirable ordering amongst groups) E andF room in the same team I fixed it however I have actually a doubt that it is wrong . Mine answer is 576 Please aid to solve this problem.
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We have actually two responses for you
Well an initial you should realize the your price can"t it is in correct together a probability is a number in between 0 and also 1.
Let"s think about E&F being in the very same group. That way that the other 4 people are separated into 2 groups. How countless ways can you perform that? You might pick a human in 4 ways and then a 2nd in 3 ways and also that would certainly seem to indicate 12 ways however picking A then B doesn"t give a various answer 보다 picking B climate A therefore we have to divide by 2, i.e. There space (4x3)/2 = 6 means in total to have actually E&F together in a group.
To uncover the probability the you desire you have to divide her answer here, namely 6, by the total variety of ways to separation the 6 human being into 3 groups. I"ll get you started on that full number. Let"s choose two civilization for the 1st group. You could pick any type of one of the 6 first and then any one that the continuing to be 5 to offer (maybe) 30 means to gain the an initial two - is that ideal or do we should divide by 2? when we"ve finished doing this calculate correctly, we understand from over we can break-up the staying 4 world into two teams in 6 ways so the total number of ways to separation 6 human being into 3 teams is (???) x 6.
Hope this it s okay you top top the right track,
A probability is always a number between zero and also 1, so solution of 576 can not be correct.
Since the six world are randomly divided into three groups of two, each such department of the six people into three teams of two is same likely. To uncover the probability the such a division involves E and also F in the same team of two, we will certainly let m = the variety of ways to division the six people into three teams of 2 so the E and F space in the same group. We will certainly let n = the total number of ways to division the six people into three teams of two. Then the answer to the difficulty will it is in m/n. Due to the fact that m and also n space both positive and m0 and 0!=1 by definition. For instance 6!=6*5*4*3*2*1=720. The 2nd idea is that the quantity "a choose b" is the variety of ways to choose b people from a team of "a" people where a>0, b>=0 and also a>=b. This is sometimes abbreviated as aCb. For example, 6C2 is the variety of ways to choose 2 people from 6 unique individuals. ACb is computed together a!/(b!(a-b)!) hence 6C2=6!/(4!2!)=15.
Let"s start by computer n i beg your pardon is the total variety of ways to division six people into three groups of two. To count the variety of ways we will divide this trouble into a sequence of tasks. The number of ways to finish the bigger trouble will it is in the product of the numbers of methods to finish each task (with one adjustment in ~ the end).
The work will be: 1. Randomly choose the first group of 2 from the six people (6C2) ways to carry out this. 2. Randomly choose the second group of two from the remaining four individuals (4C2) means to do this. 3. Randomly select the 3rd group of 2 from the remaining two people (2C2) means to carry out this.
The product that the number of means to do tasks 1,2,3 is (6C2)*(4C2)*(2C2).
However, an alert that the tasks create a "first group", "second group" and also "third group". We don"t yes, really care about the setup of the groups, for example "AB", "CD", "EF" is really the very same as the "EF", "AB", "CD". So making use of these tasks we have actually counted each division of the six people into three teams of two much more than once. Since there are 3! methods to arrange 3 pairs, then utilizing (6C2)(4C2)(2C2) counts each department of the six individuals into three teams of 2 3! times. For this reason the total number of ways to divide six human being into three teams of 2 is n=(6C2)*(4C2)*(2C2)/3! = 15
We have the right to use similar reasoning to counting the number of ways to divide the six human being into three teams of two so the E and also F room in the exact same group. 1. Realize the E and also F space in a group of two (1 way to perform this). 2. Randomly choose two world from the remaining four for the secondgroup (4C2 ways). 3. Randomly choose two people from the staying two because that the 3rd group(2C2 ways). As above m= 1*4C2*2C2/2! = 3 (EF is always in the first group)
Thus the answer come this probability concern is 3/15=1/5=0.2
A shorter way to settle this trouble is the following:
Let x it is in the number of ways to division four civilization into two groupings of two human being each.
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Then n = 6C2*x and m=1*x so the answer is 1*x/6C2*x = 1/5=0.2
Hope this is correct I"m a bit rusty!