I NEED HELP. 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 a 1 and then picking a 1? 1 2 3 4

There would be a 1/221 chance.

Step-by-step explanation:

So in a stack of cards, there are 52 total cards, 4 of which are 1's.

With this, you would have a 1/13 chance of picking up a card that's a 1 for the first picking, since 1/13 = 4/52.

Since this 1 card was picked and not put back by the second picking, the chance of picking up a card that is a one drops. With this, there is now a 3 / 51 chance that a 1 card is  picked up since this card was removed.

Now with combining these two probabilities, multiply them together:

1/13 * 3/51 = 3/663 ---> 1/221.

x = 12

y = 15

step-by-step explanation:

i only know the sequence of arithmetic it adds the first is 5+4 then it goes up as it adds to then it is 9+5 then 14+6 then to 20+7 and so on. 5,9,14,20,27,35.

step-by-step explanation: