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# 2 numbers to be selected from 0-9. What is the p that the

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Intern
Joined: 08 Sep 2004
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2 numbers to be selected from 0-9. What is the p that the [#permalink]  09 Sep 2004, 04:31
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2 numbers to be selected from 0-9. What is the p that the product of two numbers is even

thanks
Director
Joined: 16 Jun 2004
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1. Total no. of ways of choosing 2 numbers out of the 10 numbers = 10C2 = 45.

2. The choice of the two numbers shouldbe such that one of them have to be even and the other number can be any of the remaining numbers -so that the prod of tjhe numbers will be even.

We have 2, 4, 6, 8 as the even numbers. Selecting one among them is 4C1 ways = 4 ways.

Having selected one even number we can choose the other number from any of the remaining 9 numbers in 9C1 or 9 ways. Remember 0 is a even number.

The probability is thus (4*9)/45 = 4/5
Manager
Joined: 02 Apr 2004
Posts: 224
Location: Utrecht
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I have 7/9 too.

Only possibility when product is odd, when both integers are odd.
Therefore,
5/10 * 4/9 = 2/9

1 - 2/9 = 7/9

Please correct me if I am wrong.

Regards,

Alex
Intern
Joined: 08 Sep 2004
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what about the posibility of getting 0. that way number is niether + or - should we not count 0 in the mix as well
so shud ans be 1- p(of 0)+p(of odd)
Manager
Joined: 02 Apr 2004
Posts: 224
Location: Utrecht
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Kudos [?]: 14 [0], given: 0

From what I know, 0 is considered even.

Regards,

Alex
Manager
Joined: 16 May 2004
Posts: 65
Location: columbus
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Kudos [?]: 4 [0], given: 0

venksune wrote:
1. Total no. of ways of choosing 2 numbers out of the 10 numbers = 10C2 = 45.

2. The choice of the two numbers shouldbe such that one of them have to be even and the other number can be any of the remaining numbers -so that the prod of tjhe numbers will be even.

We have 2, 4, 6, 8 as the even numbers. Selecting one among them is 4C1 ways = 4 ways.

Having selected one even number we can choose the other number from any of the remaining 9 numbers in 9C1 or 9 ways. Remember 0 is a even number.

The probability is thus (4*9)/45 = 4/5

couple of things venksune

first off nowhere is it mentioned that numbers cant be repeated
so I think the total number od cases is 10* 10 = 100

another thing is you considered the case where the first number of the product is even , but what about the condition where the 2nd number is even

so I think the toatl number of favourable cases are 4*9 + 9*4 = 72

so the final ans is 72/100
Manager
Joined: 02 Apr 2004
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Location: Utrecht
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Another approach:

Total integers is 10
Total even integers = 5 (0,2,4,6,8)
Total odd integers = 5 (1,3,5,7,9)
Possible selections are:

Even and Even
Even and Odd
Odd and Even
Odd and Odd

E and E:
5/10 * 4/9 = 20/90 (product is even)
E and O:
5/10 * 5/9 = 25/90 (product is even)
O and E:
5/10 * 5/9 = 25/90 (product is even)
O and O:
5/10 * 4/9 = 20/90 (product is odd)

Therefore propability of an even product is:
90/90 - 20/90 = 70/90 --> 7/9
or
20/90 + 25/90 + 25/90 = 70/90 --> 7/9

What do you guys think?
Please correct me if I am wrong.

Regards,

Alex
Director
Joined: 31 Aug 2004
Posts: 610
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Kudos [?]: 50 [0], given: 0

I agree with Alex, the prob we are looking for is = 1 - prob that 2 picked numbers are both odd

=1 - 5C2/10C2 = 1-2/9= 7/9
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