That is false because, this type of system can have one solution, two solutions, or no solutions. Graph both equations on the same coordinate plane. Identify the point of intersection, if any.
Hope I Helped : )
Let’s say, hypothetically speaking, you chose the second marble without replacing the first marble so, events are hypothetically dependent. Events are dependent if the occurrence of one hypothetical event hypothetically does affect the likelihood that the other events occur. The probably of two or more dependent events A and B is the probability of A times the probability of B after A hypothetically occurs
P(A and B) = P(A) x P(B after A)
Choose the first marble
The total number of hypothetical marbles are, hypothetically speaking, 4 on a hypothetical basis, and there is one red marble.
P(red)=1/4
Choose the second marble
Without hypothetically replacing the hypothetical first marble, you choose the hypothetical marble, hypothetically speaking. So, the total hypothetical number of marbles are, hypothetically, 3, and there is, hypothetically, one green marble.
P(green) = 1/3
The probability of choosing red and then, hypothetically, green is:
P(red and green) = P(red) x P(green)
=1/4 x 1/3
= 1/12
P(red and green) is hypothetically equal to 1/12 on a hypothetical account.
Final hypothetical answer: 1/12
