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Given that a and b are real numbers such that -3 ≤ a ≤ 1 and -2 ≤ b ≤ 4, and values for a and b are chosen at random, what is the probability that product a * B is positive?

Apr 21, 2018

#1
+2

Alright, here we go:

For two number's product to be positive, those numbers must be both positive or both negative, but cannot be zero.

This is because zero is neither positive nor negative, and 0 * n = 0.

The first step is to find the total number of combination: (a,b).

This can be done by:

$$(1-(-3)+1)\cdot(4-(-2)+1) =5\cdot7=35.$$

There are a total of 35 different combinations.

Out of those combinations there are:

( -3 , -2 ) ; ( -3 , -1 ) ; ( -2 , -2 ) ; ( -2 , -1 ) ; ( -1 , -2 ) ; ( -1 , -1 )

( 1 , 1 ) ; ( 1 , 2 ) ; ( 1 , 3 ) ; ( 1 , 4 )

10 that work.

The probabilty of this happening is:

$$\boxed{\frac{10}{35}=\frac{2}{7}}$$

I hope this helped,

Gavin.

Apr 21, 2018

#1
+2

Alright, here we go:

For two number's product to be positive, those numbers must be both positive or both negative, but cannot be zero.

This is because zero is neither positive nor negative, and 0 * n = 0.

The first step is to find the total number of combination: (a,b).

This can be done by:

$$(1-(-3)+1)\cdot(4-(-2)+1) =5\cdot7=35.$$

There are a total of 35 different combinations.

Out of those combinations there are:

( -3 , -2 ) ; ( -3 , -1 ) ; ( -2 , -2 ) ; ( -2 , -1 ) ; ( -1 , -2 ) ; ( -1 , -1 )

( 1 , 1 ) ; ( 1 , 2 ) ; ( 1 , 3 ) ; ( 1 , 4 )

10 that work.

The probabilty of this happening is:

$$\boxed{\frac{10}{35}=\frac{2}{7}}$$

I hope this helped,

Gavin.

GYanggg Apr 21, 2018