#Probability of being in the same group twice in a row

80 messages · Page 1 of 1 (latest)

atomic swallow
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"You need to split 8 people into two equally sized groups. How many possible ways can this be done? What is the probability that you and 3 other students are placed in the same group twice? The events are independent."

For the first part, I thought to create two groups of size 4 where order does not matter (A,B,C,D = D,C,B,A) , $\frac{8!}{4!4!} = 70$, which is correct. For the last question, out of these 70 groups, only one group contains me and the other previous 3 students, hence 1/70. The answer correct answer is 1/35

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atomic swallow
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Do the probabilities add up? I,e first time me being in the group with the 3 students is 1/70, then the second time is 1/70+1/70 = 1/35? If so, when can I view that the probabilities add up if they are independent?

atomic swallow
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Bump

untold elm
atomic swallow
untold elm
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Yea

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Without replacement and order doesnt matter is binomial coeffcient

atomic swallow
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I think you are allowed to place back students into different groups, hence why I got 70 which is correct

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(8,2) = 28 right?

untold elm
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It's actually 8 over 4 mb

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which is also 70

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From 8 people we pick 4 random people

atomic swallow
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Yup

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I just don't understand why the probability of beirg in the same group twice in a row is 1/35 and not 1/70

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There's more questions after this, i.e gettign two same students, including you in a row, only one student and you in a row, and no student. All of them have 35 as a denominator

untold elm
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Gimme a sec

atomic swallow
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I've been at this for 2 hours nad 41 minutes so no rush xD

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I've essentiall cleared my whole day to just understand this problem

untold elm
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But that would make the question misleading

atomic swallow
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Wouldn't that be 50 procent chance then?

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no nono

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I don't know : /

untold elm
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Out of 70 groups we have actually 35 groups for A and 35 groups for B

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So like there are two possibilities in which group you four might end up

atomic swallow
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35 pairs?

untold elm
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yea

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35 pairs for A and B

atomic swallow
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Sorry for being slow, but why are we interested in the pairs and not the toal group combinations?

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Oh wait

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Is it because (A,B,C,D) and (E,F,G,H) are equal to (E,F,G,H), (A,B,C,D)

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So there's really only 35 unique combinations

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so 1/35

untold elm
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basically

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I thought of A,B,C,D can either be in group A or in group B

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Making it 2 possibilities of 70

untold elm
atomic swallow
untold elm
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Like whether you have A,B,C,D in group 1 or in group 2

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concludes to having 35 unique possibilities

atomic swallow
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I think I am getting the hang of it now

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Combinatorics is truly a difficult subject for me : /

untold elm
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I hated it too

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Barely passed but I passed

atomic swallow
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Now I understand why the other questions have 35 in the denominator, only the numerator left to figure out

untold elm
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But yea saying we have 35 unique groups is the key

atomic swallow
untold elm
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Yea you can decide whether 4 people go in group 1 or group 2, and by that doubling the combinations

untold elm
atomic swallow
atomic swallow
untold elm
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???

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wtf

atomic swallow
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You helped me out previously 4 times on this account and once on @west ferry

untold elm
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Oh

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Haha

atomic swallow
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I keep count because it's almost you who have answered

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Yesterday was Imobump

untold elm
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Also I recommend posting your q's in the other help channels

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Instead of here

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Chances are more helpers stumble upon you

atomic swallow
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Do you mean the follow up?

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I want to attempt solving those again after a small break

untold elm
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These I mean

atomic swallow
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Oh

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I thought those were taken

untold elm
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no the taken ones have a name

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name of the OP

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Those are nameless so free

atomic swallow
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Understood

untold elm
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these channels are more active

atomic swallow
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Will definetly do those next time

untold elm
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ye

atomic swallow
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Thanks once again, Adwnis!

untold elm
atomic swallow
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.close