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

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

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

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...

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.

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) = \(\frac{3}{6}\) = \(\frac{1}{2}\)

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
_________________

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

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