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

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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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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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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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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 [#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

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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 combinations-> 5C1 * 6C1
Combinations that gives us -ve value upon multiplication-> 5C1 * 3C1. [5C1 for M and 3C1 for T]

15/30, probability is 1/2. D
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If an integer is to be randomly selected from set M above [#permalink]
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

The product of two integers will be negative (lets call this, say [K]) when the signs are opposite for both integers.
Since set M contains negative integers, the probability of K occurring (P(K)) hinges on set T which has 3 positive integers out a total of 6 integers.

Hence P(K) = $$\frac{3}{6}$$ = $$\frac{1}{2}$$

Answer choice (D) is correct, IMO.
If an integer is to be randomly selected from set M above [#permalink]
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