Box Composition and Problem Statement

The problem involves a box containing a specific number of red and white balls. We are drawing two balls one after another without putting the first ball back. The goal is to find the likelihood that the second ball drawn is red.

• Number of red balls = 10
• Number of white balls = 12
• Total number of balls in the box initially = $10 + 12 = 22$

The drawing is done "without replacement", which means the total number of balls decreases after the first ball is drawn, and the number of balls of a specific color might also decrease, depending on the color of the first ball drawn.

Probability of Second Ball Being Red

To find the probability that the second ball drawn is red, we need to consider the possible outcomes of the first draw, as they affect the second draw:

1. The first ball drawn is red, AND the second ball drawn is red.
2. The first ball drawn is white, AND the second ball drawn is red.

The total probability of the second ball being red is the sum of the probabilities of these two mutually exclusive events.

Calculating Probabilities

Let's calculate the probability for each case:

Case 1: First Ball Red (R1) and Second Ball Red (R2)

• Probability of drawing a red ball first (P(R1)): There are 10 red balls out of 22 total.

  $\text{P(R1)} = \frac{\text{Number of red balls}}{\text{Total number of balls}} = \frac{10}{22}$

• After drawing one red ball without replacement, there are now $22 - 1 = 21$ balls left in the box. The number of red balls remaining is $10 - 1 = 9$.

  Probability of drawing a red ball second, given the first was red (P(R2|R1)):

  $\text{P(R2|R1)} = \frac{\text{Number of red balls remaining}}{\text{Total number of balls remaining}} = \frac{9}{21}$

• The probability of both events happening (R1 and R2) is the product of their probabilities:

  $\text{P(R1 and R2)} = \text{P(R1)} \times \text{P(R2|R1)} = \frac{10}{22} \times \frac{9}{21}$

  $\text{P(R1 and R2)} = \frac{90}{462}$

Case 2: First Ball White (W1) and Second Ball Red (R2)

• Probability of drawing a white ball first (P(W1)): There are 12 white balls out of 22 total.

  $\text{P(W1)} = \frac{\text{Number of white balls}}{\text{Total number of balls}} = \frac{12}{22}$

• After drawing one white ball without replacement, there are now $22 - 1 = 21$ balls left in the box. The number of red balls remaining is still 10 (since a white ball was drawn).

  Probability of drawing a red ball second, given the first was white (P(R2|W1)):

  $\text{P(R2|W1)} = \frac{\text{Number of red balls remaining}}{\text{Total number of balls remaining}} = \frac{10}{21}$

• The probability of both events happening (W1 and R2) is the product of their probabilities:

  $\text{P(W1 and R2)} = \text{P(W1)} \times \text{P(R2|W1)} = \frac{12}{22} \times \frac{10}{21}$

  $\text{P(W1 and R2)} = \frac{120}{462}$

Total Probability for Second Ball Red

The total probability that the second ball is red is the sum of the probabilities of Case 1 and Case 2:

$\text{P(Second ball is red)} = \text{P(R1 and R2)} + \text{P(W1 and R2)}$

$\text{P(Second ball is red)} = \frac{90}{462} + \frac{120}{462}$

$\text{P(Second ball is red)} = \frac{90 + 120}{462} = \frac{210}{462}$

Simplifying the Result

Now, we simplify the fraction $\frac{210}{462}$. We can find the greatest common divisor (GCD) of 210 and 462.

• Prime factorization of $210 = 2 \times 3 \times 5 \times 7$
• Prime factorization of $462 = 2 \times 3 \times 7 \times 11$
• The common prime factors are 2, 3, and 7.
• GCD = $2 \times 3 \times 7 = 42$

Divide the numerator and the denominator by the GCD:

$\frac{210 \div 42}{462 \div 42} = \frac{5}{11}$

So, the probability that the second ball drawn is red is $\frac{5}{11}$.

This detailed step-by-step calculation shows how to determine the probability for the second ball when drawing without replacement, considering all possible outcomes of the first draw.