Understanding the Probability Problem with Boxes and Balls

This question asks for the probability of drawing two balls of different colours from a box chosen randomly from two similar boxes, each containing red and black balls in different proportions. We need to use the concepts of conditional probability and the Law of Total Probability to solve this problem.

Analyzing the Contents of Each Box

We are given two similar boxes, B₁ and B₂. The number of balls of each colour depends on the box index 'i'.

• For Box B₁ (where i = 1):
  • Number of red balls = (i + 1) = (1 + 1) = 2
  • Number of black balls = (5 – i – 1) = (5 – 1 – 1) = 3
  • Total number of balls in B₁ = 2 + 3 = 5
• For Box B₂ (where i = 2):
  • Number of red balls = (i + 1) = (2 + 1) = 3
  • Number of black balls = (5 – i – 1) = (5 – 2 – 1) = 2
  • Total number of balls in B₂ = 3 + 2 = 5

Calculating Probability of Choosing a Box

Since one box is chosen at random from two similar boxes, the probability of choosing either box is equal.

• Probability of choosing Box B₁, P(B₁) = $\frac{1}{2}$
• Probability of choosing Box B₂, P(B₂) = $\frac{1}{2}$

Calculating Probability of Drawing Different Coloured Balls from Each Chosen Box

We want to find the probability of drawing two balls of different colours from the chosen box. This means drawing one red ball and one black ball. We will calculate this probability for each box.

Probability of drawing different coloured balls from Box B₁

Box B₁ has 2 red and 3 black balls (total 5). The number of ways to choose 2 balls from 5 is $\binom{5}{2}$. The number of ways to choose 1 red ball and 1 black ball is $\binom{2}{1} \times \binom{3}{1}$.

• Total ways to choose 2 balls from 5 = $\binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10$
• Ways to choose 1 red and 1 black ball = $\binom{2}{1} \times \binom{3}{1} = 2 \times 3 = 6$
• Probability of drawing different colours given Box B₁ was chosen, P(Different colours | B₁) = $\frac{\text{Ways to choose 1 red and 1 black}}{\text{Total ways to choose 2 balls}} = \frac{6}{10} = \frac{3}{5}$

Probability of drawing different coloured balls from Box B₂

Box B₂ has 3 red and 2 black balls (total 5). The number of ways to choose 2 balls from 5 is $\binom{5}{2}$. The number of ways to choose 1 red ball and 1 black ball is $\binom{3}{1} \times \binom{2}{1}$.

• Total ways to choose 2 balls from 5 = $\binom{5}{2} = 10$
• Ways to choose 1 red and 1 black ball = $\binom{3}{1} \times \binom{2}{1} = 3 \times 2 = 6$
• Probability of drawing different colours given Box B₂ was chosen, P(Different colours | B₂) = $\frac{\text{Ways to choose 1 red and 1 black}}{\text{Total ways to choose 2 balls}} = \frac{6}{10} = \frac{3}{5}$

Calculating the Total Probability

To find the overall probability of drawing two balls of different colours, we use the Law of Total Probability. This law states that the probability of an event (drawing different colours) is the sum of its probabilities under each possible condition (choosing B₁ or choosing B₂), weighted by the probability of each condition.

P(Different colours) = P(Different colours | B₁) $\times$ P(B₁) + P(Different colours | B₂) $\times$ P(B₂)

Substituting the calculated values:

P(Different colours) = $\left(\frac{3}{5} \times \frac{1}{2}\right) + \left(\frac{3}{5} \times \frac{1}{2}\right)$

P(Different colours) = $\frac{3}{10} + \frac{3}{10}$

P(Different colours) = $\frac{6}{10}$

P(Different colours) = $\frac{3}{5}$

Summary of Probability Calculation

The probability that both balls drawn are of different colours is $\frac{3}{5}$.

Revision Table: Key Probability Concepts

Understanding basic probability concepts is crucial for solving problems involving random events.

• Sample Space: The set of all possible outcomes of a random experiment. In this case, the sample space involves choosing a box AND drawing two balls.
• Event: A subset of the sample space. Here, the event is "drawing two balls of different colours".
• Combinations: Used to count the number of ways to choose items from a set without regard to the order. The formula is $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.
• Conditional Probability: The probability of an event occurring given that another event has already occurred. Denoted as P(A | B).
• Law of Total Probability: A fundamental theorem that relates marginal probabilities to conditional probabilities. If E is an event and A₁, A₂, ..., A_n are mutually exclusive and exhaustive events, then P(E) = $\sum_{i=1}^{n} P(E | A_i) P(A_i)$.

Additional Information: Similar Probability Problems

Problems involving drawing balls from boxes are common in probability. They often test your understanding of:

• Basic probability calculation.
• Combinations or permutations (depending on whether order matters).
• Conditional probability.
• Bayes' theorem (if the question asks for the probability of having chosen a specific box given the outcome of the ball drawing).
• Expected value (if there are costs or rewards associated with drawing certain balls).

Always carefully identify the total number of possible outcomes and the number of favourable outcomes for the specific event you are interested in. When multiple stages are involved, like choosing a box first and then drawing balls, break down the problem into conditional probabilities for each stage and combine them using appropriate rules like the Law of Total Probability.