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Probability question

by priyasaibaba » Thu Apr 22, 2010 5:11 am
Can someone help me solve this problem?

If 2 numbers are to be chosen at random from 1, 2, 3, ...10, without replacement, what is the probability that the product of two selected numbers is divisible by 10?
(A) 1/5
(B) 1/3
(C) 4/15
(D) 13/45
(E) 14/45

Thanks,
Priya
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by ajith » Thu Apr 22, 2010 5:22 am
priyasaibaba wrote:Can someone help me solve this problem?

If 2 numbers are to be chosen at random from 1, 2, 3, ...10, without replacement, what is the probability that the product of two selected numbers is divisible by 10?
(A) 1/5
(B) 1/3
(C) 4/15
(D) 13/45
(E) 14/45

Thanks,
Priya
There are two ways to do this

a. Select 5 and select any even number other than 10
b. Select 10 and select any number

These are mutually exclusive

a. can be done in 1*4 = 4 ways
b can be done 1*9 = 9 ways

Two numbers can be selected in 10*9/2 = 45 ways

The probability = 13/45
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by analyst218 » Thu Apr 22, 2010 9:55 am
ajith wrote:
priyasaibaba wrote:Can someone help me solve this problem?

If 2 numbers are to be chosen at random from 1, 2, 3, ...10, without replacement, what is the probability that the product of two selected numbers is divisible by 10?
(A) 1/5
(B) 1/3
(C) 4/15
(D) 13/45
(E) 14/45

Thanks,
Priya
There are two ways to do this

a. Select 5 and select any even number other than 10
b. Select 10 and select any number

These are mutually exclusive

a. can be done in 1*4 = 4 ways
b can be done 1*9 = 9 ways

Two numbers can be selected in 10*9/2 = 45 ways

The probability = 13/45
why would you leave out 10x5 in your (a)?
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by meghash3 » Thu Apr 22, 2010 10:29 am
got it til the point we say 45 ways of selecting the number, how is probability 13/45 ?? pl elaborate
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by analyst218 » Thu Apr 22, 2010 10:59 am
meghash3 wrote:got it til the point we say 45 ways of selecting the number, how is probability 13/45 ?? pl elaborate
well i got 14/45 not 13/45.
in order to be divisible by 10, the two numbers being multiplied together
has to have either 5 or 10 in it.

1. any even number times 5 ends with a 0 which is divisible by 10.
my reasoning is there are 5 cases since there's 5 even number from 1 to 10 inclusive.

2. any integer times 10 will be a multiple of 10 so 9 cases. since you're not replacing
10.
so total 5+9=14

14/45.

ajith is a smart guy so i may have gotten it wrong. still waitin on why he didnt include 5x10
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by harshavardhanc » Thu Apr 22, 2010 11:31 am
priyasaibaba wrote:Can someone help me solve this problem?

If 2 numbers are to be chosen at random from 1, 2, 3, ...10, without replacement, what is the probability that the product of two selected numbers is divisible by 10?
(A) 1/5
(B) 1/3
(C) 4/15
(D) 13/45
(E) 14/45

Thanks,
Priya
take any even number with 5 (5 cases)

OR

any number with 10 ( 9 cases)

therefore, reqd. prob = (9+5)/ 10C2 = 14/45.
Regards,
Harsha
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by priyasaibaba » Thu Apr 22, 2010 12:12 pm
Hi,

The OA is 13/45.

Analyst218, the reason for not including 5*10 in the first case is because it is included in the 2nd case.

Thanks,
Priya
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by harshavardhanc » Thu Apr 22, 2010 12:53 pm
priyasaibaba wrote:Hi,

The OA is 13/45.

Analyst218, the reason for not including 5*10 in the first case is because it is included in the 2nd case.

Thanks,
Priya
missed that part.... it's my second mistake today in PS ! :(
Regards,
Harsha
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by frank1 » Sat Apr 24, 2010 6:45 am
Well,
i got 13/45
but with the method
10=9
9-0
8-1....and so on...
which was much manual and lenthy ....

I dont know if it applies for other situation or not
but seems like
10-9
5-4
total 13 was perfect shortcut for this

thanks
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