Answer:
![4.82\cdot 10^{-4}](https://tex.z-dn.net/?f=4.82%5Ccdot%2010%5E%7B-4%7D)
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:
![p(a)=\frac{a}{n}=\frac{4}{52}=\frac{1}{13}=0.0769](https://tex.z-dn.net/?f=p%28a%29%3D%5Cfrac%7Ba%7D%7Bn%7D%3D%5Cfrac%7B4%7D%7B52%7D%3D%5Cfrac%7B1%7D%7B13%7D%3D0.0769)
At the second drawing, the ace is not replaced within the deck. So the number of cards left in the deck is
![n-1=51](https://tex.z-dn.net/?f=n-1%3D51)
Therefore, the probability of drawing a three at the 2nd draw is
![p(t)=\frac{t}{n-1}=\frac{4}{51}=0.0784](https://tex.z-dn.net/?f=p%28t%29%3D%5Cfrac%7Bt%7D%7Bn-1%7D%3D%5Cfrac%7B4%7D%7B51%7D%3D0.0784)
Then, at the third draw, the previous 2 cards are not replaced, so there are now
![n-2=50](https://tex.z-dn.net/?f=n-2%3D50)
cards in the deck. So, the probability of drawing a jack is
![p(j)=\frac{j}{n-2}=\frac{4}{50}=0.08](https://tex.z-dn.net/?f=p%28j%29%3D%5Cfrac%7Bj%7D%7Bn-2%7D%3D%5Cfrac%7B4%7D%7B50%7D%3D0.08)
Therefore, the total probability of drawing an ace, a three and then a jack is:
![p(atj)=p(a)\cdot p(j) \cdot p(t)=0.0769\cdot 0.0784 \cdot 0.08 =4.82\cdot 10^{-4}](https://tex.z-dn.net/?f=p%28atj%29%3Dp%28a%29%5Ccdot%20p%28j%29%20%5Ccdot%20p%28t%29%3D0.0769%5Ccdot%200.0784%20%5Ccdot%200.08%20%3D4.82%5Ccdot%2010%5E%7B-4%7D)