Question

Find the probability that three successive face cards are drawn in three successive draws (without replacement) from a deck of cards. Define Events \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\) as follows: Event A: a face card is drawn on the first draw, Event \(B\) : a face card is drawn on the second draw, Event \(C\) : a face card is drawn on the third draw.

Short answer

The probability of drawing three successive face cards without replacement is \(\frac{11}{221}\).

Step-by-step solution

Find the probability of Event A

Event A: a face card is drawn on the first draw. We have 12 face cards and 52 cards in total, so the probability of event A happening is: \[P(A) = \frac{12}{52} = \frac{3}{13}\]

Find the probability of Event B given Event A

Event B: a face card is drawn on the second draw, given that a face card was already drawn in the first draw. Now, there are 11 face cards remaining and 51 cards in total. The probability of event B given event A is: \[P(B|A) = \frac{11}{51}\]

Find the probability of Event C given Events A and B

Event C: a face card is drawn on the third draw, given that face cards were drawn in the first two draws. There are now 10 face cards left and 50 cards in total. The probability of event C given events A and B is: \[P(C|A, B) = \frac{10}{50} = \frac{1}{5}\]

Find the joint probability

Now, we need to find the probability of all three events A, B, and C happening. We can do this by multiplying the probabilities found in steps 1, 2, and 3: \[P(A \cap B \cap C) = P(A) \times P(B|A) \times P(C|A, B) = \frac{3}{13} \times \frac{11}{51} \times \frac{1}{5}\]

Calculate the probability

Lastly, we plug in the fractions to find the probability of drawing three successive face cards without replacement: \[P(A \cap B \cap C) = \frac{3}{13} \times \frac{11}{51} \times \frac{1}{5} = \frac{33}{663} = \frac{11}{221}\] So, the probability of drawing three successive face cards without replacement is \(\frac{11}{221}\).

Key concepts

Conditional Probability

Understanding the likelihood of an event in relation to another event is at the heart of conditional probability. It addresses the question: 'What is the probability of event B occurring, given that event A has already happened?' The notation for conditional probability is usually expressed as \( P(B|A) \), which reads as 'the probability of B given A'.

In the context of the exercise, the calculation of the probability that a face card is drawn second, given that a face card has already been drawn first (Event B given Event A), illustrates this concept. As the events are sequential without replacement, the sample space is reduced with each draw, affecting the probability of subsequent events. The calculated probabilities in the step-by-step solution, such as \( P(B|A) = \frac{11}{51} \), exemplify the idea of adjusting our probability assessments based on new information - a cornerstone of conditional probability.

Joint Probability

Joint probability is a statistical measure that calculates the likelihood of two or more events happening at the same time. It's represented by the notation \( P(A \cap B) \), which signifies the probability of event A and event B occurring together. This concept extends to any number of events, with the joint probability of events A, B, and C, for example, represented as \( P(A \cap B \cap C) \).

In the exercise, we are interested in the joint probability of drawing three successive face cards (Events A, B, and C). The step-by-step solution provides the joint probability by multiplying the independent probabilities of each event: \( P(A) \), \( P(B|A) \), and \( P(C|A, B) \). This multiplication rule only holds true because the events are dependent; the occurrence of each preceding event alters the probability of the succeeding event.

Probability Without Replacement

The concept of probability without replacement is a critical part of understanding how probabilities change in sequential events. When we remove an item from a dataset and don't return it, like drawing a card from a deck and not putting it back, the composition of the dataset changes with each draw. This impacts the likelihood of a specific outcome on the next draw.

The exercise perfectly illustrates this concept. Each card drawn changes the makeup of the deck, thus altering the probabilities of drawing another face card in subsequent draws. The reduction in the number of face cards and total cards available for drawing ensures that each event is influenced by the previous ones, and the solution shows the step-by-step recalibration of probabilities after each draw.