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.
_________________

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

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

P= No of ways the product of two integers is negative/All possible values
= 5*3/5*6 =1/2

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

Probability = 15 / 30 = 1/2

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

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

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
_________________

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.

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

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...

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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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.

This question appears to have changed in the web version...and it appears broken:

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

Out of interest, would any one have solved this by setting up a multiplication table on their pad? (i.e. see below)

x (-6) (-5) (-4) (-3) (-2)
(-2) + + + + +
(-1) + + + + +
(0) 0 0 0 0 0
(1) - - - - -
(2) - - - - -
(3) - - - - -

Total negative = 15
Total options = 30

--> 15/30 = 1/2