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@Apu
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+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.@SirLancelotDuLac
Either each element is in P or in in Q or in P intersection Q so 3^n cases for this thingy.
Total number of cases for P and Q is 2^n.2^n toh 4^n
Hence probability is (0.75)^n
Oh okk
And also the 3rd partplz
In these kinds of questions (combinatorially) make 4 boxes. In P, In Q, In both P&Q and in neither
Noted
Choose k elements from n, then you can have 2^k stuff for Q (Q can be phi)
So the total number of cases is sum of nCk2^k
which is 3^k
So same answer
The answer is 3n/4^n
Not 3^n
Think of it this way
n=0 hota
toh original set phi hota
So P and Q must be phi
Yes
since phi is a proper subset of phi
The prob should have been 1
which is (0.75)^n when n=0
What
We talkin bout the same quesn right?
Yes
Ab smjha
Yeah Q must be proper subset of P wala na?
Yea i got it now
Wait the one marked as 4th or one marked as 3rd?
3rd
Crap sorry
Above is for 4th one
Tbbhi sochu subsets kyu aare ha
3rd ke soln mein
Third wale ke liye tohh similar approach
ek element chuno (nC1) woh teen dabbe mein jaa sakta hai
Either P only, Q only or P and Q both
Baaki neither mein
So 3*nC1 cases
divided by 4^n total cases
so 3n/4^n
Should it also be 3^n
Ek element pe 3 choices ha
Nahi
We have to choose one element and then only that element has 3 choices
Baaki sabh already fixated rehte hain neither wale box mein
Acha pehle sab elements pe 3 chocie thi
Idhar ek ha
Isliye 3n
Right?
Yep.
Alright
Thanx
Np 🫡
Solved mark karu aapko, vrna retire hojaoge💀
Release me ðŸ˜
Prolly just 2 more to go after these open threads lol
Abhi to bilkul nahiðŸ˜
+solved @SirLancelotDuLac
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