#probability

Question

8 pairs of shoes are in a closet. 8 shoes are selected at random. The probability that there will be at least one pair and at most 3 pairs amongst selected shoes are?

Comments

1 - ((8C4+2⁸)/16C8) = 0.9746

@papesh is this correct??? I'm not sure...

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good approach

Answer 1

1st case:we have taken 1 pair.then from 8 pairs,we can choose it in 8 ways.now we have to take another 6 shoes from remaining 7 pairs, which will not be pair of anyone.this will be ⁷C₆ * 2⁶ =7*64=448(choosing 6 pairs from 7. after that taking any one of them from each pair).so total ways=(8*448)=3584

2nd case:we have taken 2 pairs.then from 8 pairs,we can choose it in ⁸C₂=28 ways.now we have to take another 4 shoes from remaining 6 pairs, which will not be pair of anyone.this will be ⁶C₄ * 2⁴ =15*16=240(choosing 4 pairs from 6. after that taking any one of them from each pair).so total ways=(28*240)=6720

3rd case:we have taken 3 pairs.then from 8 pairs,we can choose it in ⁸C₃=56 ways.now we have to take another 2 shoes from remaining 5 pairs, which will not be pair of anyone.this will be ⁵C₂ * 2²=10*4=40(choosing 2 pairs from 5. after that taking any one of them from each pair).so total ways=(56*40)=2240

so required probability=(3584+6720+2240)/¹⁶C₈=6272/6435=0.9746

Answer 2

S = 16C8;
A = (2C2x14C6)+(4C4x12C4)+(6C6x10C2) = 3543;
P = A/S = 0.275.
Explanation: Firstly think the probability has to come very less to 1,beacause der r 16 shoes and the pairs r being randomly chosen,so definitely 0.947 is not the answer!LOL!
Now just see der r 16 shoes(8 pairs) ::::
The meaning of 8 shoes randomly selected is,u can choose either 4 pairs or in random way
Now atleast 1 and atmost 3 pairs means, among 16 u can choose 2 shoes(which has to be a pair)
so 2C2,and remaining 6 shoes r randomly selected from 14,so the event for atleast one is::::2C2x14C6
I think this is simple question! and d rest of all is the same way!!!

Comments

@shailesh11

I think there is a flaw in this logic because in this term 14C6 u are selecting any 6 shoes from 14 shoes. Now in this selection we may have 1more pair or 2 more pairs or 3 more pairs. Also you are counting them thrice because of terms 12C4 and 10C2.

Also i don't know why u find the answer 0.9746 funny. In selecting 8 shoes from 16 shoes chances of getting no pair and chances of getting all 4 pairs is very low.16C8 is 12870.In case of no pair, Possiblities of getting no pair are 2^8=256 because from each pair we can select only one shoes i.e. either left or right but not both, 8 such pairs are there.Hence, 2^8. Fraction 256/12870 is 0.019 which is very low. Similarly, to get all 4 pairs, we have to select 4 pairs from 8 given shoes pairs. So 8C4=70 which is 70/12870.

Subtracting the sum of these probabilities from 1 gives 0.9746

Hence the answer of atleast one and atmost three is closer to 1.

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Sorry!,Ive solved it in 5 mins,didnt notice that in 14 there might be pairs can come out,but after asserting this logic,i might get the answer.Thanks for reply.

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can anyone suggest me that how would i can improve my probabality >?????????????????????

give me some links plz

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http://gatecse.in/probability/

http://gatecse.in/combinatory/

Specially IITM link for counting is very helpful in when to use what formula and imagining situation in each problem in terms of balls and bins

For various probability distributions, gatecse link is very helpful.

For conditional probability

https://youtu.be/JGeTcRfKgBo

This MIT video will clear all ur concepts

Hope it helps