Answer:
D. If John owns a dog, then he owns a cat
Step-by-step explanation:
The implication p → q (if p, then q) has the same truth table as the logical expression ~p∨q. You have the expression ...
~(John owns a dog) ∨ (he owns a cat)
Matching parts of this expression to the components of the expression ~p∨q, we see we can choose ...
- p = John owns a dog
- q = he owns a cat
and put those into the structure of the implication: if p, then q.
If John owns a dog, then he owns a cat. . . . . matches choice D
Answer:
The probability is 0.0052
Step-by-step explanation:
Let's call A the event that the four cards are aces, B the event that at least three are aces. So, the probability P(A/B) that all four are aces given that at least three are aces is calculated as:
P(A/B) = P(A∩B)/P(B)
The probability P(B) that at least three are aces is the sum of the following probabilities:
- The four card are aces: This is one hand from the 270,725 differents sets of four cards, so the probability is 1/270,725
- There are exactly 3 aces: we need to calculated how many hands have exactly 3 aces, so we are going to calculate de number of combinations or ways in which we can select k elements from a group of n elements. This can be calculated as:

So, the number of ways to select exactly 3 aces is:

Because we are going to select 3 aces from the 4 in the poker deck and we are going to select 1 card from the 48 that aren't aces. So the probability in this case is 192/270,725
Then, the probability P(B) that at least three are aces is:

On the other hand the probability P(A∩B) that the four cards are aces and at least three are aces is equal to the probability that the four card are aces, so:
P(A∩B) = 1/270,725
Finally, the probability P(A/B) that all four are aces given that at least three are aces is:
