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If an integer is to be randomly selected from set M above

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If an integer is to be randomly selected from set M above [#permalink]

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M = {-6, -5, -4, -3, -2}
T = {-2, -1, 0, 1, 2, 3}

If an integer is to be randomly selected from set M above and an integer is to be randomly selected from set T above, what is the probability that the product of the two integers will be negative?

A. 0
B. 1/3
C. 2/5
D. 1/2
E. 3/5
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Re: good prob [#permalink]

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New post 12 Jan 2008, 04:40
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marcodonzelli wrote:
M = {-6, -5, -4, -3, -2}
T = {-2, -1, 0, 1, 2, 3}

If an integer is to be randomly selected from set M above and an integer is to be randomly selected from set T above, what is the probability that the product of the two integers will be negative?

A. 0
B. 1/3
C. 2/5
D. 1/2
E. 3/5


I think D is correct
in order to get negative integer either m should be negative and n positive
probability of m is negative =5/5=1
probability of n is positive 3/6=1/2
probability of m negative and m positive is 1*1/2=1/2
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Re: good prob [#permalink]

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marcodonzelli wrote:
M = {-6, -5, -4, -3, -2}
T = {-2, -1, 0, 1, 2, 3}

If an integer is to be randomly selected from set M above and an integer is to be randomly selected from set T above, what is the probability that the product of the two integers will be negative?

A. 0
B. 1/3
C. 2/5
D. 1/2
E. 3/5


P= No of ways the product of two integers is negative/All possible values
= 5*3/5*6 =1/2
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Re: good prob [#permalink]

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M = {-6, -5, -4, -3, -2}
T = {-2, -1, 0, 1, 2, 3}

If an integer is to be randomly selected from set M above and an integer is to be randomly selected from set T above, what is the probability that the product of the two integers will be negative?

A. 0
B. 1/3
C. 2/5
D. 1/2
E. 3/5

Soln: Total number of possible ways of choosing two integers is = 5 * 6 = 30 ways
Now for the product of two integers to be chosen to be negative = they should be of opposite sign. Since set M has all negative numbers, thus we move to set T which has 3 positive numbers.
Thus total number of possible ways in which product will be negative is = 5 * 3 = 15

Probability that the product of the two integers will be negative
= 15/30
= 1/2
Ans is D
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Re: good prob [#permalink]

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New post 14 Feb 2010, 14:27
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marcodonzelli wrote:
M = {-6, -5, -4, -3, -2}
T = {-2, -1, 0, 1, 2, 3}

If an integer is to be randomly selected from set M above and an integer is to be randomly selected from set T above, what is the probability that the product of the two integers will be negative?

A. 0
B. 1/3
C. 2/5
D. 1/2
E. 3/5


Negative prod pairs = 5 x 3 = 15
total pairs possible = 5c1 x 6c1 = 5 x 6 = 30

Probability = 15 / 30 = 1/2
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Re: good prob [#permalink]

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New post 14 Feb 2010, 23:00
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If product has to be negative then m is -ve and n is +ve or m is +ve and n is -ve. But all elements of m are -ve hence we need n to be +ve.
0 has to be excluded as any number multiplied by 0 is 0 and it is neither +ve nor -ve.

p(m&n) = p(m)*p(n) = 1*3/6 = 1/2 hence D
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Re: If an integer is to be randomly selected from set M above and an integ [#permalink]

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New post 27 Aug 2010, 22:49
Answer D.
Total # of outcomes: 5*6= 30
# of outcomes where Product is -ve : (-6,1) ,(-6,2), (-6,3)... Hence, total: 15
Probability: 15/30 = 1/2
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Re: If an integer is to be randomly selected from set M above and an integ [#permalink]

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New post 14 Sep 2010, 01:46
udaymathapati wrote:
M = {-6, -5, -4, -3, -2}
T = {-2, -1, 0, 1, 2, 3}
If an integer is to be randomly selected from set M above and an integer is to be randomly selected from set T above, what is the probability that the product of the two integers will be negative?
A. 0
B. 1/3
C. 2/5
D. 1/2
E. 3/5



Please explain where my logic is going wrong...

Required probablity=probability of selecting a -ve no from 1st set * probability of selecting a +ve no from 2st set excluding (0)

answer=(1/5) * (3/6)=1/10

which is none o fteh choices!!! Please let me know my mistake...
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Re: If an integer is to be randomly selected from set M above and an integ [#permalink]

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New post 14 Sep 2010, 01:48
utin wrote:
udaymathapati wrote:
M = {-6, -5, -4, -3, -2}
T = {-2, -1, 0, 1, 2, 3}
If an integer is to be randomly selected from set M above and an integer is to be randomly selected from set T above, what is the probability that the product of the two integers will be negative?
A. 0
B. 1/3
C. 2/5
D. 1/2
E. 3/5



Please explain where my logic is going wrong...

Required probablity=probability of selecting a -ve no from 1st set * probability of selecting a +ve no from 2st set excluding (0)

answer=(1/5) * (3/6)=1/10

which is none o fteh choices!!! Please let me know my mistake...


All numbers in set M are negative
Probability of selecting a negative number = 1
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Re: M = {-6, -5, -4, -3, -2} T = {-2, -1, 0, 1, 2, 3} If an [#permalink]

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New post 30 Mar 2012, 23:25
Hello,

I got a doubt. Any no. mulitplied by 0 is 0 and 0 is neither +ve nor -ve. Going by this prop of 0, y shld we consider 0 as one of the nos. cause any no +ve or -ve multiplied by 0 is neither positive nor negative ?

Thnx.
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Re: M = {-6, -5, -4, -3, -2} T = {-2, -1, 0, 1, 2, 3} If an [#permalink]

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New post 31 Mar 2012, 00:34
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priyalr wrote:
Hello,

I got a doubt. Any no. mulitplied by 0 is 0 and 0 is neither +ve nor -ve. Going by this prop of 0, y shld we consider 0 as one of the nos. cause any no +ve or -ve multiplied by 0 is neither positive nor negative ?

Thnx.


It's not clear what you mean by "why should we consider 0 as one of the numbers"... Anyway:

M = {-6, -5, -4, -3, -2}
T = {-2, -1, 0, 1, 2, 3}
If an integer is to be randomly selected from set M above and an integer is to be randomly selected from set T above, what is the probability that the product of the two integers will be negative?

A. 0
B. 1/3
C. 2/5
D. 1/2
E. 3/5

In order the product of two multiples to be negative they must have different signs. Since Set M consists of only negative numbers then in order mt to be negative we should select positive number from set T, the probability of that event is 3/6=1/2, (since out of 6 number in the set 3 are positive).

Answer: D.

Hope it's clear.
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Re: If an integer is to be randomly selected from set M above [#permalink]

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marcodonzelli wrote:
M = {-6, -5, -4, -3, -2}
T = {-2, -1, 0, 1, 2, 3}

If an integer is to be randomly selected from set M above and an integer is to be randomly selected from set T above, what is the probability that the product of the two integers will be negative?

A. 0
B. 1/3
C. 2/5
D. 1/2
E. 3/5


Total number of cases = 6*6
To get a negative result, we need to multiply numbers of negative signs.
Set M has all the 6 negative numbers.
Set T has 3 positive numbers, which when multiplied by negative numbers will yield a negative number

Hence we have 6*3 favourable cases.

Probability = 18/36 = 1/2
Option D
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Re: If an integer is to be randomly selected from set M above [#permalink]

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New post 14 Mar 2017, 19:07
kbulse wrote:
marcodonzelli wrote:
M = {-6, -5, -4, -3, -2}
T = {-2, -1, 0, 1, 2, 3}

If an integer is to be randomly selected from set M above and an integer is to be randomly selected from set T above, what is the probability that the product of the two integers will be negative?

A. 0
B. 1/3
C. 2/5
D. 1/2
E. 3/5


I think D is correct
in order to get negative integer either m should be negative and n positive
probability of m is negative =5/5=1
probability of n is positive 3/6=1/2
probability of m negative and m positive is 1*1/2=1/2


Set M has all negative numbers, so, the probability of selecting a negative number lies entirely on set M.
Set M has 2 negative, 1 zero and 3 positive numbers. Thus, P(negative product) = \(\frac{3}{6}\) = \(\frac{1}{2}\)
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Re: If an integer is to be randomly selected from set M above [#permalink]

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New post 15 Mar 2017, 09:13
Shruti0805 wrote:
kbulse wrote:
marcodonzelli wrote:
M = {-6, -5, -4, -3, -2}
T = {-2, -1, 0, 1, 2, 3}

If an integer is to be randomly selected from set M above and an integer is to be randomly selected from set T above, what is the probability that the product of the two integers will be negative?

A. 0
B. 1/3
C. 2/5
D. 1/2
E. 3/5


I think D is correct
in order to get negative integer either m should be negative and n positive
probability of m is negative =5/5=1
probability of n is positive 3/6=1/2
probability of m negative and m positive is 1*1/2=1/2


Set M has all negative numbers, so, the probability of selecting a negative number lies entirely on set M.
Set M has 2 negative, 1 zero and 3 positive numbers. Thus, P(negative product) = \(\frac{3}{6}\) = \(\frac{1}{2}\)


This question would get much more tricky if 2/3 was one of the altenatives...
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Re: If an integer is to be randomly selected from set M above [#permalink]

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marcodonzelli wrote:
M = {-6, -5, -4, -3, -2}
T = {-2, -1, 0, 1, 2, 3}

If an integer is to be randomly selected from set M above and an integer is to be randomly selected from set T above, what is the probability that the product of the two integers will be negative?

A. 0
B. 1/3
C. 2/5
D. 1/2
E. 3/5


In order for the product of the two integers to be negative, one of them has to be negative and the other has to be positive. Since every integer in set M is negative, we must select a positive integer in set T.
Thus, the probability of selecting a negative number in set M and then a positive number in set T is:

5/5 x 3/6 = 1/2

Answer: D
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Re: If an integer is to be randomly selected from set M above [#permalink]

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New post 14 Apr 2017, 12:25
marcodonzelli wrote:
M = {-6, -5, -4, -3, -2}
T = {-2, -1, 0, 1, 2, 3}

If an integer is to be randomly selected from set M above and an integer is to be randomly selected from set T above, what is the probability that the product of the two integers will be negative?

A. 0
B. 1/3
C. 2/5
D. 1/2
E. 3/5


Any number picked from the first set will have the same properties (all -ve).
A positive number is required from the second set to make the product -ve. 1/2 chance of picking a +ve number from the second set.
Re: If an integer is to be randomly selected from set M above   [#permalink] 14 Apr 2017, 12:25
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