Two cards are drawn at random from a shuffled deck. (a) What is the probability that at least one is a heart? (b) If you know that at least one is a heart, what is the probability that both are hearts?

Combinatorial probability is a key area of probability theory that deals with counting the number of ways certain events can occur. These counts are often described using a combination formula. In our exercise, we used combinatorial probability to figure out the number of possible ways to draw 2 cards from a deck of 52 cards. The combination formula is given as \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \] where \(n\) is the total number of items, and \(k\) is the number of items to choose. By applying this formula, we find \[ \binom{52}{2} = \frac{52!}{2!(52-2)!} = \frac{52 \times 51}{2 \times 1} = 1326 \] which is the total number of ways to pick 2 cards from 52. Similarly, for non-hearts and hearts cases, we calculated combinations like \[ \binom{39}{2} \] and \[ \binom{13}{2} \] to further understand the different outcomes possible.

Conditional probability deals with the probability of an event occurring given that another event has already occurred. It is denoted as \( P(A | B) \), which reads as 'the probability of A given B.' In our exercise, we considered the probability of both cards being hearts given that at least one card drawn is a heart. To determine this, we followed a step-by-step calculation:

• First, we calculated the number of ways to have at least one heart, subtracting non-heart combinations from the total combinations: \[ P(\text{At Least One Heart}) = 1326 - 741 = 585 \]
• Next, we calculated the probability of both cards being hearts if at least one is known to be a heart: \[ P(\text{Both Hearts} | \text{At Least One Heart}) = \frac{\binom{13}{2}}{585} = \frac{78}{585} \]
This conditional probability helps us understand how additional information affects the likelihood of outcomes.

Card probabilities are specific applications of combinatorial and conditional probability. When dealing with cards, common tasks include calculating the chances of drawing certain types or suits of cards in various scenarios. In our problem, we addressed two key questions:

• What is the probability of drawing at least one heart out of two cards? By calculating \[ P(\text{No Hearts}) = \frac{\binom{39}{2}}{\binom{52}{2}} = \frac{741}{1326} \] and its complement, we found that \[ P(\text{At Least One Heart}) = 1 - P(\text{No Hearts}) = 1 - \frac{741}{1326} = \frac{585}{1326} \]
• What is the chance that both cards are hearts, given at least one is a heart? We determined that: \[ P(\text{Both Hearts} | \text{At Least One Heart}) = \frac{\binom{13}{2}}{585} = \frac{78}{585} \]

These calculations underscore the interconnectedness of different probability concepts and their specific application to card scenarios, making it easy to grasp the underlying patterns in card probability problems.