Question

There is a box containing 5 white balls, 4 black balls, and 7 red balls. If two balls are drawn one at a time from the box and neither is replaced, find the probability that (1) both balls will be white. (2) the first ball will be white and the second red. (3) if a third ball is drawn, find the probability that the three balls will be drawn in the order white, black, red.

Short answer

The probabilities for the given events are as follows: 1. Probability of both balls being white: \( \frac{1}{12} \) 2. Probability of first white, then red: \( \frac{35}{240} \) 3. Probability of the order white, black, red: \( \frac{1}{28} \)

Step-by-step solution

Event 1: Both Balls Will Be White

First, let's find the probability that both drawn balls are white. To do this, we need to count the number of successful outcomes (both balls are white) and divide it by the total number of possible outcomes (all ways to draw 2 balls from the box). Total number of balls = 5 white + 4 black + 7 red = 16 balls First ball drawn: Number of successful outcomes (white ball) = 5 Total number of possible outcomes = 16 So, the probability of drawing a white ball first is \( \frac{5}{16} \). Second ball drawn: Since one white ball is already drawn and not replaced, we now have 4 white balls left. Number of successful outcomes (white ball) = 4 Total number of possible outcomes = 15 (as one ball is already drawn) So, the probability of drawing a white ball second is \( \frac{4}{15} \). In order to find the probability of both events happening together, we multiply the individual probabilities: Event 1 probability = \( \frac{5}{16} \cdot \frac{4}{15} = \frac{20}{240} = \frac{1}{12} \)

Event 2: First White, Then Red

Now, let's find the probability that the first ball will be white and the second red. We will again compute the probabilities of individual events and then multiply them. First ball drawn (white ball): Probability = \( \frac{5}{16} \) Second ball drawn (red ball): Number of successful outcomes (red ball) = 7 Total number of possible outcomes = 15 (as one ball is already drawn) So, the probability of drawing a red ball second is \( \frac{7}{15} \). Event 2 probability = \( \frac{5}{16} \cdot \frac{7}{15} = \frac{35}{240} \)

Event 3: White, Black, Red in Order

Finally, let's find the probability that the three balls will be drawn in the order white, black, red. This time we need to find the probabilities of three individual events and multiply them. First ball drawn (white ball): Probability = \( \frac{5}{16} \) Second ball drawn (black ball): Number of successful outcomes (black ball) = 4 Total number of possible outcomes = 15 (as one ball is already drawn) So, the probability of drawing a black ball second is \( \frac{4}{15} \). Third ball drawn (red ball): Since one white and one black ball are already drawn and not replaced, we now have 6 red balls left. Number of successful outcomes (red ball) = 6 Total number of possible outcomes = 14 (as two balls are already drawn) So, the probability of drawing a red ball third is \( \frac{6}{14} \). Event 3 probability = \( \frac{5}{16} \cdot \frac{4}{15} \cdot \frac{6}{14} = \frac{120}{3360} = \frac{1}{28} \) Summarizing our results: 1. Probability of both balls being white: \( \frac{1}{12} \) 2. Probability of first white, then red: \( \frac{35}{240} \) 3. Probability of the order white, black, red: \( \frac{1}{28} \)

Key concepts

Combinatorics

Combinatorics is a branch of mathematics that deals with counting, arranging, and calculating possibilities. In probability, we often use combinatorics to figure out how many ways events can occur.
For example, in this exercise, you have a box containing different colored balls. You're asked to figure out the likelihood of drawing balls in certain orders. Combinatorics helps us here by allowing us to:

• Count the total number of ways to select two or more balls.
• Identify the number of successful or favorable outcomes for specific conditions like drawing both white or particular sequences like white followed by red.

To find these probabilities, we typically use the concept of permutations when the order is important, or combinations when it is not. In this exercise, the order does matter, so we focus on permutations.

Conditional Probability

Conditional probability is the probability of an event occurring given that another event has already happened. It is key when dealing with problems where the initial conditions or available outcomes change after each event, like in this exercise where balls are drawn without replacement.
When calculating the probability of sequential events, such as drawing a white ball followed by a red ball, we use the formula for conditional probability:

• \[ P(B \mid A) = \frac{P(A \cap B)}{P(A)} \]

This expression reads as "the probability of event B occurring given that event A has occurred". Here, the sequence of draws affects the probability calculations since each successive event's probability depends on the outcomes of the previous draws.
This method ensures you take into account the changing numbers of balls as they are drawn without replacement, allowing for accurate probability assessments.

Dependent Events

Dependent events are events where the outcome or occurrence of one event affects the likelihood of another event. In this exercise, each draw from the box is a dependent event because removing a ball changes the probabilities for the next draw.
Here's how this works in practice:

• When you draw the first white ball, the total number of balls decreases, as does the count of white balls, which affects the odds of drawing another white one next.
• Similarly, after drawing a white and a black ball, you have different probabilities for drawing a red one next, because the total count and available colors have changed.

Understanding how events affect each other is crucial in probability. With dependent events, probability calculations must consider prior outcomes to correctly calculate subsequent probabilities. This ensures you accurately determine the chances of complex events, such as sequences like white, black, and then red in order.