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If an integer is to be randomly selected from set M above [#permalink]
 12 Jan 2008, 05:29
Question Stats:
59% (01:43) correct
40% (00:52) wrong based on 126 sessions
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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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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Joined: 07 Nov 2007
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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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Manager
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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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Senior Manager
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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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Senior Manager
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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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All numbers in the first bank are negative so it has to select a positive from the second bank, but zero won't work since it would be zero.
5/5 * 3/6 = 15/30 => 1/2
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Re: M = {-6, -5, -4, -3, -2} T = {-2, -1, 0, 1, 2, 3} If an [#permalink]
31 Mar 2012, 00:25
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]
31 Mar 2012, 01:34
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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: M = {-6, -5, -4, -3, -2} T = {-2, -1, 0, 1, 2, 3} If an [#permalink]
05 Oct 2012, 22:26
Bunuel wrote: 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. Why us 0 not considered as the product of any number will give non negetive
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Re: M = {-6, -5, -4, -3, -2} T = {-2, -1, 0, 1, 2, 3} If an
[#permalink]
05 Oct 2012, 22:26
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