Answer:

Step-by-step explanation:
In a deck of cart, we have:
a = 4 (aces)
t = 4 (three)
j = 4 (jacks)
And the total number of cards in the deck is
n = 52
So, the probability of drawing an ace as first cart is:

At the second drawing, the ace is not replaced within the deck. So the number of cards left in the deck is

Therefore, the probability of drawing a three at the 2nd draw is

Then, at the third draw, the previous 2 cards are not replaced, so there are now

cards in the deck. So, the probability of drawing a jack is

Therefore, the total probability of drawing an ace, a three and then a jack is:
