Please help! I've been on this question for more than an hour... You pick a card at random. Without putting the first card back, you pick a second card at random.

What is the probability of picking an even number and then picking an even number?

Simplify your answer and write it as a fraction or whole number.

1/6

Step-by-step explanation:

Out of these four card choices, for the first pick there are two even cards and four cards in total. This means that on the first pick there is a 2/4=1/2 chance that you pick an even card. On the second pick, if you do not replace the card, then there is 1 even card remaining, and 3 cards in total, leaving a probability of 1/3. Multiplying these two probabilities together, you get an overall chance of 1/6. Hope this helps!

step-by-step explanation:

ok so im not 100% sure but you would start by adding 5.1t and 1.9t beacuse they have the same variable. that is 7t.   then add -3.6 + 1.2 which is -2.4

so then keep that as -2.4 + 7t. i apologize if this is incorrect. best luck on your test < 3