Math Problem Statement

There are 2n black counters and 2n + 1 green counters in a bag. If two counters are randomly drawn, one after the other, show that the probability of getting two counters the same colour is \(\frac{2n}{n + 1}\).

Solution

To solve this problem, let’s define the components and calculate the probability of drawing two counters of the same color.

  1. Counters in the Bag:

    • The total number of black counters is 2n2n.
    • The total number of green counters is 2n+12n + 1.
    • The total number of counters is 2n+(2n+1)=4n+12n + (2n + 1) = 4n + 1.
  2. Possible Outcomes: We want to calculate the probability of drawing two counters of the same color. This can happen in two cases:

    • Both counters are black.
    • Both counters are green.
  3. Probability of Drawing Two Black Counters:

    • The probability of drawing the first black counter is 2n4n+1\frac{2n}{4n+1}.
    • After one black counter is drawn, there are 2n12n - 1 black counters left out of a total of 4n4n counters.
    • Thus, the probability of drawing a second black counter is 2n14n\frac{2n - 1}{4n}.
    • Therefore, the probability of drawing two black counters is: 2n4n+1×2n14n=2n(2n1)(4n+1)4n\frac{2n}{4n+1} \times \frac{2n - 1}{4n} = \frac{2n \cdot (2n - 1)}{(4n + 1) \cdot 4n}
  4. Probability of Drawing Two Green Counters:

    • The probability of drawing the first green counter is 2n+14n+1\frac{2n + 1}{4n + 1}.
    • After one green counter is drawn, there are 2n2n green counters left out of a total of 4n4n counters.
    • Thus, the probability of drawing a second green counter is 2n4n\frac{2n}{4n}.
    • Therefore, the probability of drawing two green counters is: 2n+14n+1×2n4n=(2n+1)2n(4n+1)4n\frac{2n + 1}{4n + 1} \times \frac{2n}{4n} = \frac{(2n + 1) \cdot 2n}{(4n + 1) \cdot 4n}
  5. Total Probability of Drawing Two Counters of the Same Color: Now, we add the probabilities of the two cases (both black or both green): 2n(2n1)(4n+1)4n+(2n+1)2n(4n+1)4n=2n(2n1)+(2n+1)2n(4n+1)4n\frac{2n \cdot (2n - 1)}{(4n + 1) \cdot 4n} + \frac{(2n + 1) \cdot 2n}{(4n + 1) \cdot 4n} = \frac{2n (2n - 1) + (2n + 1) \cdot 2n}{(4n + 1) \cdot 4n} Simplifying the numerator: =4n22n+4n2+2n(4n+1)4n=8n2(4n+1)4n= \frac{4n^2 - 2n + 4n^2 + 2n}{(4n + 1) \cdot 4n} = \frac{8n^2}{(4n + 1) \cdot 4n} Simplify further: =2nn+1= \frac{2n}{n + 1}

Thus, the probability of drawing two counters of the same color is indeed 2nn+1\frac{2n}{n + 1}, as required.


Would you like further clarification on any steps?

Related Questions:

  1. What would the probability be if there were an odd number of black counters?
  2. How does the probability change if we draw with replacement?
  3. What if there were 2n2n green counters instead of 2n+12n + 1?
  4. How can we calculate the probability for more than two counters drawn?
  5. Can we generalize this probability formula for any two colors?

Tip:

When working with probabilities, identify and separate each possible case, then combine the cases carefully.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Probability
Combinatorics
Independent and Dependent Events

Formulas

Probability of drawing two black counters: \(\frac{2n}{4n+1} \times \frac{2n - 1}{4n}\)
Probability of drawing two green counters: \(\frac{2n + 1}{4n + 1} \times \frac{2n}{4n}\)

Theorems

Addition Rule for Probabilities
Multiplication Rule for Independent Events

Suitable Grade Level

Grades 11-12