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Algebra

New kid on the Block

 Joined: 25 Mar 2012 Post: 28
3 Apr 2012 10:34:25 IST
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Engineering Entrance , JEE Main , JEE Main & Advanced , Mathematics , Algebra

Fifteen coupons are numbered 1,2,........,15 respectively.Seven coupons are selected at random one at a time with replacement.The probability that the largest number appearing on a selected coupon is 9,is

Blazing goIITian

Joined: 27 Dec 2010
Posts: 1508
3 Apr 2012 10:44:47 IST
1 people liked this

number of ways to select 7 coupons = C(15,7).

we select 9.

the rest 6 to be selected from 1 - 8.

ways = C(8,6).

probabilty is C(8,6) / C(15,7) ???

galat ho sakta hai. probability nahi kari acche se..

Blazing goIITian

Joined: 2 Mar 2012
Posts: 308
3 Apr 2012 11:06:26 IST
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[9^7-8^7]/[15^7]

New kid on the Block

Joined: 25 Mar 2012
Posts: 28
3 Apr 2012 11:11:47 IST
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Blazing goIITian

Joined: 21 Mar 2012
Posts: 1457
3 Apr 2012 16:52:17 IST
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it is with replacement so n(s)=15^7 now...........we have to select n(E) such that 9 maximum appears so every ball must be selected from first none balls we have n(E)=9^7 so p(E)=9^7/15^7

New kid on the Block

Joined: 25 Mar 2012
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4 Apr 2012 08:44:14 IST
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New kid on the Block

Joined: 4 Apr 2012
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4 Apr 2012 09:02:35 IST
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9^7/15^7 is the anser

New kid on the Block

Joined: 4 Apr 2012
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4 Apr 2012 09:04:39 IST
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ory puka nuvuu keka

Blazing goIITian

Joined: 27 Dec 2010
Posts: 1508
4 Apr 2012 09:09:11 IST
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i didn't read the word - "replacement"!!!

shit...

Blazing goIITian

Joined: 27 Dec 2010
Posts: 1508
4 Apr 2012 09:22:38 IST
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so now i know what it means,

number of ways to pick up 7 coupons is 15^7. cause we have 15 choices on each time we pick.

there 9^7 ways to pick the coupons numbered 1 to 9.

we want at least one 9.

therefore number of ways = (total ways) - (ways where we don't have 9 and select from 1 to 8) = 9^7 - 8^7.

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