Question

A box contains 5 red balls and 4 green balls. Two balls are drawn one after another without replacement. What is the probability that the second ball is green, given that the first ball drawn was red?

• A. \(\dfrac{1}{2}\)
• B. \(\dfrac{5}{18}\)
• C. \(\dfrac{2}{5}\)
• D. \(\dfrac{4}{8}\)

Hint:

Use conditional probability formula: \[ P(B \mid A) = \frac{P(A \cap B)}{P(A)} \] and adjust the total count for “without replacement.”

Answer 1

The Correct Option is A

To solve the problem of finding the probability that the second ball is green, given that the first ball drawn was red, we need to process the events and their probabilities.

Step 1: Determine Total Initial Outcomes
The box initially contains 5 red balls and 4 green balls, so a total of 9 balls.

Step 2: Probability of First Ball Being Red
The probability of drawing a red ball first is given by:

\[ P(\text{Red first}) = \frac{5}{9} \]

Step 3: Probability of Second Ball Being Green Given First Is Red
After drawing one red ball, there are now 8 balls left in the box. Out of these, 4 are green. The probability that the second ball drawn is green, given the first was red, is:

\[ P(\text{Green second | Red first}) = \frac{4}{8} = \frac{1}{2} \]

Therefore, the probability that the second ball drawn is green, given that the first ball drawn was red, is \(\frac{1}{2}\).

Answer 2

We are given:
- Total red balls = 5
- Total green balls = 4
- Total balls = 9
The event is:
- First ball drawn is red
- Second ball is green
Since the first red ball is already drawn, we now have:
- 4 red balls
- 4 green balls
- Remaining total = 8
So the conditional probability is:
\[ P(\text{2nd is green} \mid \text{1st is red}) = \frac{\text{Favorable green balls}}{\text{Remaining total balls}} = \frac{4}{8} = \boxed{\dfrac{1}{2}} \]