We have to calculate the probability of picking a 4 and then a 5 without replacement.
We can express this as the product of the probabilities of two events:
• The probability of picking a 4
,
• The probability of picking a 5, given that a 4 has been retired from the deck.
We have one card in the deck out of fouor cards that is a "4".
Then, the probability of picking a "4" will be:
![P(4)=\frac{1}{4}](https://tex.z-dn.net/?f=P%284%29%3D%5Cfrac%7B1%7D%7B4%7D)
The probability of picking a "5" will be now equal to one card (the number of 5's in the deck) divided by the number of remaining cards (3 cards):
![P(5|4)=\frac{1}{3}](https://tex.z-dn.net/?f=P%285%7C4%29%3D%5Cfrac%7B1%7D%7B3%7D)
We then calculate the probabilities of this two events happening in sequence as:
![\begin{gathered} P(4,5)=P(4)\cdot P(5|4) \\ P(4,5)=\frac{1}{4}\cdot\frac{1}{3}=\frac{1}{12} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20P%284%2C5%29%3DP%284%29%5Ccdot%20P%285%7C4%29%20%5C%5C%20P%284%2C5%29%3D%5Cfrac%7B1%7D%7B4%7D%5Ccdot%5Cfrac%7B1%7D%7B3%7D%3D%5Cfrac%7B1%7D%7B12%7D%20%5Cend%7Bgathered%7D)
Answer: 1/12