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A bag contains ten balls numbered 1 to 10

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A bag contains ten balls numbered 1 to 10 [#permalink] New post 28 Nov 2019, 00:00
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Question Stats:

45% (00:46) correct 54% (00:33) wrong based on 11 sessions
A bag contains ten balls numbered 1 to 10. If 2 balls are selected from the bag with replacement,what is the probability that at least one of them is an even numbered ball

(a)1/2
(b)1/3
(c)3/4
(d)3/5
(e)4/5
[Reveal] Spoiler: OA

Last edited by Carcass on 28 Nov 2019, 00:05, edited 1 time in total.
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Re: A bag contains ten balls numbered 1 to 10 [#permalink] New post 28 Nov 2019, 00:06
Expert's post
P(At least one of them is even numbered) = 1- p( Both of them are odd)

= 1- 5/10 = 5/10 =3/4 or 0.75
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Re: A bag contains ten balls numbered 1 to 10 [#permalink] New post 03 Dec 2019, 12:30
total draws: 10*10 = 100
two odds: 5*5 = 25
P = 25/100= 1/4
atleast 1 even(=1-p): 1-1/4 --> 3/4

Answer: C
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Re: A bag contains ten balls numbered 1 to 10 [#permalink] New post 06 Dec 2019, 03:43
Carcass wrote:
P(At least one of them is even numbered) = 1- p( Both of them are odd)

= 1- 5/10 = 5/10 =3/4 or 0.75


Shouldn't it be 1-(5c2/10c2) =1-(10/45)=7/9 ?
Re: A bag contains ten balls numbered 1 to 10   [#permalink] 06 Dec 2019, 03:43
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A bag contains ten balls numbered 1 to 10

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