Question

There are 10 white and 5 black balls in a bag. Two balls are drawn one by one. First ball is not placed back before the second is taken out. Assume that the taking out of each ball from the bag is equally likely. What is the probability that both balls taken out are white?

Hint:

For "without replacement" problems, remember to adjust both the number of favorable outcomes and the total number of outcomes for each subsequent draw. An alternative method is using combinations: \(P = \frac{\binom{10}{2}}{\binom{15}{2}}\), which gives the same result.

Answer 1

Step 1: Understanding the Concept:
This is a problem of conditional probability, specifically drawing without replacement. The outcome of the second draw depends on the outcome of the first draw because the first ball is not returned to the bag.
Step 2: Key Formula or Approach:
The probability of two dependent events A and B occurring in sequence is given by the multiplication rule of probability:
\(P(\text{A and B}) = P(A) \times P(B|A)\)
Where \(P(A)\) is the probability of the first event, and \(P(B|A)\) is the probability of the second event given that the first event has occurred.
Let A be the event that the first ball is white, and B be the event that the second ball is white.
Step 3: Detailed Explanation or Calculation:
First, identify the initial state of the bag:
Number of white balls = 10
Number of black balls = 5
Total number of balls = 10 + 5 = 15
Event A: The first ball drawn is white.
The probability of drawing a white ball first is: \[ P(A) = \frac{\text{Number of white balls}}{\text{Total number of balls}} = \frac{10}{15} \] Event B|A: The second ball drawn is white, given the first was white.
After drawing one white ball without replacement, the state of the bag changes:
Number of white balls remaining = 10 - 1 = 9
Total number of balls remaining = 15 - 1 = 14
The probability of drawing a second white ball is: \[ P(B|A) = \frac{\text{Number of remaining white balls}}{\text{Total number of remaining balls}} = \frac{9}{14} \] Now, we find the probability that both balls are white: \[ P(\text{Both white}) = P(A) \times P(B|A) \] \[ P(\text{Both white}) = \frac{10}{15} \times \frac{9}{14} \] Simplify the fractions: \[ P(\text{Both white}) = \frac{2}{3} \times \frac{9}{14} = \frac{18}{42} \] Reduce the fraction to its simplest form by dividing the numerator and denominator by their greatest common divisor, which is 6: \[ P(\text{Both white}) = \frac{18 \div 6}{42 \div 6} = \frac{3}{7} \] Step 4: Final Answer:
The probability that both balls taken out are white is \(\frac{3}{7}\).