Conditional Probability and Independence

Conditional Probability and Dependence

Imagine drawing a card from a deck before a friend draws the next one. Does your card change the probability of what your friend can draw? Without replacement, it often does. Conditional probability calculates the second probability using what is already known.

In a finite sample space with equally likely outcomes, the conditional probability of event $B$ given event $A$ is the fraction of outcomes in $A$ that also belong to $B$. If conditioning on $A$ changes the probability of $B$, the events are dependent. If the probability stays the same, they are independent.

Before calculating the next probability, restrict the possible outcomes using the condition that is already known.

Mathematical Formula and Notation

For events $A$ and $B$ with $P(A) > 0$, conditional probability is defined by:

The quantities in this formula are:

• $P(B|A)$ is the probability of $B$ under the condition that $A$ occurred
• $P(A \cap B)$ is the probability that $A$ and $B$ both occur
• $P(A)$ is the probability of the conditioning event $A$

Rearranging the definition gives the general multiplication rule:

If the events are independent, learning $A$ does not change the probability of $B$:

Application to Card Drawing

Drawing without replacement changes the sample space after each card leaves the deck. Both examples below update the probabilities at the second draw. Because the first card is not returned, the denominator of the second probability is one smaller.

Ace and King Scenario

A standard deck has $52 \text{ cards}$. It contains $4 \text{ Aces}$ and $4 \text{ Kings}$. Two cards are drawn without replacement. What is the probability of drawing an Ace first and a King second?

Step-by-step analysis:

Let:

• Event $C$ is drawing an Ace first
• Event $D$ is drawing a King second

Probability calculation:

After one Ace is drawn, $51 \text{ cards}$ remain. All $4 \text{ Kings}$ are still present, so the conditional probability of drawing a King is $\frac{4}{51}$.

Probability of Drawing Two Hearts in Sequence

To see how the favorable count can change, consider drawing two Hearts in a row:

• First draw: $P(A) = \frac{13}{52} = \frac{1}{4}$
• Second draw after getting a heart: $P(B|A) = \frac{12}{51}$

The probability of drawing Hearts on both draws is therefore:

Application in Management Selection

A company reviews the distribution of managers by gender and job level. Its data has the following composition:

Let $L$ denote the event that a selected manager is male.

The company randomly selects two managers without replacement for a committee. The probability that both selected managers are male is:

After selecting one male manager, there are $599$ total managers remaining with $199$ male managers, so the probability of selecting a second male manager decreases to $\frac{199}{599}$.

How to Update the Sample Space Before Multiplying

The first step below replaces the original sample space with the outcomes that remain. The later steps show how that replacement separates a conditional probability from an independent product.

Update the Sample Space Before Multiplying

To solve conditional probability problems:

1. Identify the events first and second clearly
2. Determine the conditions after the first event occurs
3. Calculate the probability of the first event from initial conditions
4. Calculate the conditional probability of the second event after the first event
5. Multiply both probabilities to get the joint probability

Difference from Ordinary Probability

For independent events, the condition does not change the next probability. For dependent events, the known outcome changes either the favorable count, the total count, or both.

Concrete example: drawing two Aces

Suppose we want to calculate the probability of getting two aces in succession from a standard deck ($52 \text{ cards}$, $4$ aces). Let $A_1$ denote the event that the first card is an ace and $A_2$ the event that the second card is an ace.

1. Independent Probability (With Replacement):

   If after drawing the first card, the card is returned and the deck is reshuffled:

   $$P(A_1) = \frac{4}{52} = \frac{1}{13}$$

   $$P(A_2) = \frac{4}{52} = \frac{1}{13}$$

   $$\begin{aligned} P(A_1 \cap A_2) &= \frac{4}{52} \times \frac{4}{52} \\ &= \frac{1}{13} \times \frac{1}{13} \\ &= \frac{1}{169} \end{aligned}$$

2. Dependent Draw (Without Replacement):

   If the first card is not returned:

   $$P(A_1) = \frac{4}{52} = \frac{1}{13}$$

   $$P(A_2|A_1) = \frac{3}{51} = \frac{1}{17}$$

   $$\begin{aligned} P(A_1 \cap A_2) &= \frac{4}{52} \times \frac{3}{51} \\ &= \frac{1}{13} \times \frac{1}{17} \\ &= \frac{1}{221} \end{aligned}$$

Result Comparison:

• Independent probability: $\frac{1}{169} = 0.00592$
• Conditional probability: $\frac{1}{221} = 0.00452$

In this example, drawing without replacement gives a smaller result because the number of remaining Aces decreases from $4$ to $3$, while the total cards decrease from $52$ to $51$.

Exercises

1. A box contains $8 \text{ red balls}$ and $12 \text{ blue balls}$. If two balls are drawn sequentially without replacement, calculate the probability of getting a red ball on the first draw and a blue ball on the second draw.

2. A deck of cards is shuffled randomly. Three cards are drawn successively without replacement. What is the probability of getting an Ace on the first draw, a King on the second draw, and a Queen on the third draw?

Try each exercise before reading its worked solution. At every draw, write both the favorable count and the new total before multiplying.

Worked Solutions

1. Solution:

   Let $A$ be getting a red ball first, and let $B$ be getting a blue ball second

   Total initial balls is $8 + 12 = 20 \text{ balls}$

   $$P(A) = \frac{8}{20} = \frac{2}{5}$$

   $$P(B|A) = \frac{12}{19}$$

   $$\begin{aligned} P(A \cap B) &= \frac{8}{20} \times \frac{12}{19} \\ &= \frac{2}{5} \times \frac{12}{19} \\ &= \frac{24}{95} \end{aligned}$$

   After drawing one red ball, there are $19 \text{ balls}$ remaining in total ($8 - 1 = 7$ red, $12$ blue). The number of blue balls does not change (remains $12$), so the probability of drawing a blue ball second is $\frac{12}{19}$.

2. Solution:

   Let $A$ be getting an Ace first, let $K$ be getting a King second, and let $Q$ be getting a Queen third

   A standard deck has $52 \text{ cards}$, including $4$ Aces, $4$ Kings, and $4$ Queens.

   $$P(A) = \frac{4}{52} = \frac{1}{13}$$

   $$P(K|A) = \frac{4}{51}$$

   $$P(Q|A \cap K) = \frac{4}{50} = \frac{2}{25}$$

   $$\begin{aligned} P(A \cap K \cap Q) &= \frac{4}{52} \times \frac{4}{51} \times \frac{4}{50} \\ &= \frac{1}{13} \times \frac{4}{51} \times \frac{2}{25} \\ &= \frac{8}{16575} \end{aligned}$$

   Each draw reduces the total number of cards ($52$→$51$→$50$). The next required rank still has all $4$ cards available: drawing an Ace does not remove a King, and drawing a King does not remove a Queen.