Question

A bag contains 7 red and 7 black coloured balls. A person drawn two balls from the bag, what is the probability that the two balls are the same in colour? (1) \(\frac{6}{13}\) (2) \(\frac{2}{7}\) (3) \(\frac{4}{13}\) (4) \(\frac{1}{13}\)

Short answer

Answer: The probability that two balls drawn from the bag have the same color is 6/13.

Step-by-step solution

Identify the Probabilities for Each Color

First, we need to find out the probability that both balls are red and the probability that both balls are black. The probability of drawing a red ball on the first draw is 7/14 (because there are 7 red balls and 14 total balls in the bag), and the probability of drawing a red ball on the second draw, given that the first ball was red, is 6/13 (because now there are only 6 red balls and 13 total balls remaining in the bag). Similarly, the probability of drawing a black ball on the first draw is 7/14, and the probability of drawing a black ball on the second draw, given that the first ball was black, is 6/13.

Calculate the Joint Probabilities

Next, we will find the joint probability of drawing two red balls and two black balls. The joint probability of drawing two red balls is the product of the probabilities of drawing a red ball on the first draw and drawing a red ball on the second draw, given that the first ball was red: \((7/14) \times (6/13) = \frac{42}{182}\). Similarly, the joint probability of drawing two black balls is the product of the probabilities of drawing a black ball on the first draw and drawing a black ball on the second draw, given that the first ball was black: \((7/14) \times (6/13) = \frac{42}{182}\).

Add the Joint Probabilities

Finally, we will add the joint probabilities of drawing two red balls and two black balls to calculate the probability that the two balls are of the same color. The probability of drawing two balls of the same color is the sum of the joint probabilities of drawing two red balls and two black balls: \(\frac{42}{182} + \frac{42}{182} = \frac{84}{182}\).

Simplify the Fraction

The last step is to simplify the fraction resulting from the previous step: \(\frac{84}{182}\) can be simplified to \(\frac{6}{13}\). Therefore, the probability that the two balls are of the same color is \(\frac{6}{13}\), which corresponds to option (1).

Key concepts

Joint Probability

Joint probability refers to the likelihood of two different events happening at the same time. In probability problems like the one given, where we draw two balls from a bag, we are interested in finding the combined chance for success during both draws.

To calculate joint probability, you usually multiply the individual probabilities of each event. For example, if you want the probability of two red balls being picked, you calculate the probability of picking a red ball first, and then multiply it by the probability of picking another red ball, assuming you don't put the first one back.

• First, calculate the probability of drawing a red ball: \( \frac{7}{14} \), because there are 7 red balls from a total of 14.
• Then, the probability of drawing another red ball becomes \( \frac{6}{13} \), as one red ball is removed from both the total and the reds.
• Multiply these probabilities for the joint probability of both: \((\frac{7}{14}) \times (\frac{6}{13}) = \frac{42}{182}\).

This step is repeated for drawing two black balls, given they also start at 7 and likewise reduce by one after a draw.

Simplifying Fractions

Simplifying fractions is an essential skill in mathematics, especially when dealing with probability problems. It helps in reducing fractions to their simplest form, making them easier to understand and work with.

In our problem, you calculate your probabilities, and they end up as fractions. The goal is to simplify them for clarity and easier comparison or general usability. Here's a straightforward method for simplifying the fraction \(\frac{84}{182}\):

• Find the greatest common divisor (GCD) of the numerator and the denominator. In this case, the GCD of 84 and 182 is 14.
• Divide both the numerator and the denominator by 14: \(\frac{84}{182} = \frac{6}{13}\).

After these steps, your fraction becomes simplified, showing clearly that \(\frac{6}{13}\) is the probability of drawing two balls of the same color.

Combinatorics

Combinatorics deals with counting, arranging, and combining objects. It is extremely useful in probability as it helps to calculate the number of possible outcomes or ways an event can occur.

In this problem, combinatorics is applied indirectly. Essentially, it is calculating the different ways you can draw balls from a set. Given you have two colors with identical counts, knowing the number of total combinations is a step toward computing probabilities. If needed, combinations can be calculated using the formula: \[\binom{n}{r} = \frac{n!}{r! (n-r)!}\]where \(n\) is the total number of objects to choose from, and \(r\) is the number of objects to choose. The factorial symbol \(!\) denotes multiplying a sequence descending natural numbers.

• For our problem: calculate possible two-ball combinations: \(\binom{14}{2}\) since we pick 2 out of 14 balls.
• No explicit calculation is made in the provided step-by-step, but this underlies how probabilities get determined, by considering all possibilities.

By comprehending combinatorics, you grasp the basis of why certain probabilities add up to their given values.

Basic Probability Concepts

Understanding basic probability concepts is crucial for solving any probability problem. Probability helps quantify the odds of an event occurring, expressed as a fraction or percentage between 0 (impossibility) and 1 (certainty).

The fundamental rule in probability is to consider all possible outcomes and the favorable outcomes. The probability is calculated with the formula: \[ P(event) = \frac{\text{number of favorable outcomes}}{\text{total number of outcomes}} \]

• Probability ranges from 0 to 1, or 0% to 100%.
• A simple event, like flipping a coin, offers a clear illustration (\(P(\text{heads}) = \frac{1}{2}\)).
• In the given problem, think of probabilities as assigning a measure to every potential outcome (similar-colored balls in this example).

Among the occurrences, discrening between mutually exclusive outcomes (where one event does not affect another) and calculating total probability via addition and multiplication rules is key. Master these concepts to become proficient in understanding randomness.