Two Red Balls from Bag (No Replacement)
A bag contains 5 red balls and 3 blue balls. If you draw 2 balls from the bag without replacement, what is the probability that both balls drawn are red?
Related Concepts
Hint
Remember that "without replacement" means the total number of balls and the number of balls of a specific color change after each draw.
- Sequential Approach:
- Probability of the 1st ball being red?
- Given the 1st was red, what's the new count of red balls and total balls for the 2nd draw?
- Combinatorial Approach:
- How many ways to choose 2 red balls from the available red balls?
- How many ways to choose any 2 balls from the total number of balls?
Explanation: Probability of Drawing Two Red Balls
Imagine a bag with 5 red balls and 3 blue balls (total 8 balls). You're going to pick two balls, one after the other, without putting the first one back.
What's the chance both are red?
- First Pick: There are 5 red balls out of 8 total. So, the chance of picking a red ball first is 5/8.
- Second Pick (after picking a red first): Now, imagine you successfully picked a red ball. There are only 4 red balls left, and only 7 total balls left in the bag. So, the chance of the second ball also being red is 4/7.
To get the probability of both these things happening, we multiply these chances: (5/8) × (4/7).
The bag initially contains 5 red balls and 3 blue balls, making a total of 5 + 3 = 8 balls. We are drawing 2 balls without replacement.
Method 1: Sequential Probability (Conditional Probability)
Let R₁ be the event that the first ball drawn is red, and R₂ be the event that the second ball drawn is red.
- Probability of the first ball being red (P(R₁)):
There are 5 red balls out of 8 total balls.
P(R₁) = 5/8 - Probability of the second ball being red, given the first was red (P(R₂|R₁)):
If the first ball drawn was red and not replaced, there are now 4 red balls left and a total of 7 balls remaining.
P(R₂|R₁) = 4/7 - Probability of both balls being red:
Using the multiplication rule:P(R₁ and R₂) = P(R₁) × P(R₂|R₁)
P(two red balls) = (5/8) × (4/7)
P(two red balls) = 20 / 56
Simplifying the fraction:20 / 56 = 5 / 14(dividing numerator and denominator by 4)
Method 2: Using Combinations
This method considers the number of ways to choose 2 red balls out of the 5 available, divided by the total number of ways to choose any 2 balls from the 8 available.
- Number of ways to choose 2 red balls from 5 red balls (C(5,2)):
C(5, 2) = 5! / (2! × (5-2)!) = 5! / (2! × 3!) = (5 × 4) / (2 × 1) = 20 / 2 = 10
There are 10 ways to choose 2 red balls. - Total number of ways to choose any 2 balls from the 8 total balls (C(8,2)):
C(8, 2) = 8! / (2! × (8-2)!) = 8! / (2! × 6!) = (8 × 7) / (2 × 1) = 56 / 2 = 28
There are 28 ways to choose any 2 balls. - Probability of drawing two red balls:
P(two red balls) = (Ways to choose 2 red balls) / (Total ways to choose 2 balls)
P(two red balls) = C(5, 2) / C(8, 2) = 10 / 28
Simplifying the fraction:10 / 28 = 5 / 14(dividing numerator and denominator by 2)
Final Result
Both methods yield the same probability:
P(drawing two red balls without replacement) = 5/14
As a decimal, this is approximately:
5/14 ≈ 0.3571 or about 35.71%
This problem highlights how probabilities change when sampling without replacement, making the events dependent. Both the sequential conditional probability approach and the combinatorial approach are valid for solving this type of problem.
Challenge: Using the same bag, what's the probability of drawing one red ball and one blue ball (in any order) without replacement?