21but 21 - 11 pair has already been counted once!!
wen v chkd 4 21 !!!
same is da case wid : 22,23,24 n 25
this is provided we cant differentiate in wich order the nos. r chosen
2 nos. are randomly selected frm 1,2,3,..........25
Wat is the prob. that Diff. b/w First and the Second no. selected is NOT less than 10
I m getting 2/5
Ans given : 1/5
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24 Answers
2121Oh okie...
so sample space ≠25C2
hmmmmmm............ point yaar!!!
k k..... 600 YEAH!!!
1Here there is no replacement, but the numbers are called first and second so they are picked seperately in 600 ways
@Amit
Yeah thats right
1ohhh 25C1 then 24 numbers are remaining for selecting second number so 24C1
so total possibilities is 600
thanks kalyan
1for the number 25
25-15>10
25-14>10
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25-1>10
so for the number 25 there are 15 possibilities
for the number 24 there are 14 possibilities
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for the number 11 there is 1 possibilitiy
thus required outcomes is 15+14+........+1=120
Taken from Amits post........ No tym to type
And the total possible outcomes is 25C124C1=600
So the required probability is 120/600=1/5[7][7]
Isnt that the solution
62yes amit had it been without replacement, my answer would have bene
240/600
but i think the question has not specifiied..
anyways we have discussed this question for both cases...
21In context to the EDITED post #3
Then it means
1) u remov a no.
2) Note its value
3) Replace it
4)Chus 2nd no.
5)note it value
then apply |x-y| >= 10
I thot the questn says sumthin else.....
21If u consider this : "for the number 11 there is 1 probabilitiy>???
isnt it 6 ways?? 1 or 21 or 22 or 23 or 24 or 25"
Each case will hav to be counted twice............
and wat abt the Denomikator then [7] [7] CONFUSED!!1
1ya sir for 11 the single probability is 11-1=10 which is not less than 10 else all possibilities like 11-2,11-3,11-4 are all less than 10
62for the number 11 there is 1 probabilitiy>???
isnt it 6 ways?? 1 or 21 or 22 or 23 or 24 or 25
one more issue.. i was selecting with replacement!!
1@ that is best method when the numbers are samll, and it is fast too[3]
62Wat is the prob. that Diff. b/w First and the Second no. selected is NOT less than 10
no it is only given difference
difference is always modulus of A-B
1fo the number 25
25-15>10
25-14>10
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.
.
..
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.
..
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.
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25-1>10
so for the number 25 there are 15 possibilities
for the number 24 there are 14 probabilities
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..
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.
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for the number 11 there is 1 probabilitiy
thus poosiblie outcomes is 15+14+........+1=120
and total number of ways of selecting two numbers is 25C2=300
p=120/300=2/5
62x, y
|x-y|>10
One big mistake I edited that..
1/25(15/25)+1/25(14/25)..........(1/25)(6/25)+(10 times)
1/25(6/25)+1/25(6/25)..........(1/25)(6/25)+(5 times)
1/25(15/25)+1/25(14/25)..........(1/25)(6/25)+(10 times)
= 210+30 = 240 ways
total no of ways is 625