Understanding the Probability of Drawing One White and One Red Ball from a Bag

Imagine a scenario where a bag contains 8 white and 3 red balls. We wish to determine the probability of drawing one white and one red ball in a specific order. This solution works even if the balls are drawn without replacement, with replacement, or under the condition that either both balls are white or both are red.

Without Replacement

When the first ball is drawn, the probability of it being white is:

Probability of first ball being white: 8/11

After drawing the first white ball, there are now 7 balls left in the bag, with 3 of them being red. Hence, the probability of the second ball being red is:

Probability of second ball being red: 3/10

The probability of drawing one white and one red ball in this order is:

Pone white, one red (without replacement) 8/11 × 3/10 24/110 12/55

With Replacement

Let's now consider the scenario where the balls are drawn with replacement. In this situation, the probability of each draw remains the same.

In the first draw, the probability of picking a white ball is:

Probability of first ball being white: 8/11

In the second draw, the probability of picking a red ball is:

Probability of second ball being red: 3/11

The probability of drawing one white and one red ball (with order considered) is:

Pone white, one red (with replacement) 8/11 × 3/11 24/121

Additional Scenarios

Additionally, let's explore the probability of drawing balls in different scenarios.

Scenario A: Both Balls the Same Color

The probability that both balls drawn are of the same color can be calculated by considering both cases where both balls are white or both are red.

For both balls to be white:

The number of ways of choosing 2 white balls out of 8 is:

8C2 8! / (2! × (8-2)!) 28

The total number of ways to choose 2 balls out of 11 is:

11C2 11! / (2! × (11-2)!) 55

The probability that both balls are white is:

Pboth white 28/55

For both balls to be red:

The number of ways of choosing 2 red balls out of 3 is:

3C2 3! / (2! × (3-2)!) 3

The total number of ways to choose 2 balls out of 11 is still 55 (as calculated earlier).

The probability that both balls are red is:

Pboth red 3/55

The combined probability that both balls are of the same color is:

Pboth same color Pboth white Pboth red 28/55 3/55 31/55

Scenario B: Order Irrelevant

In the initial example, we calculated the probability of drawing one white and one red ball in a specific order. However, considering the order irrelevant, the total probability changes.

The total number of ways to draw two balls from the bag is:

11C2 55

The number of ways to draw one white and one red ball is:

8 (ways to choose a white ball) × 3 (ways to choose a red ball) 24

Hence, the probability of drawing one white and one red ball, regardless of order, is:

Pone white and one red (order irrelevant) 24/55

Conclusion

In summary, we have explored multiple scenarios to understand the probability of drawing one white and one red ball from a bag. Whether the balls are drawn with or without replacement, the overall probability can be calculated using basic combinatorial principles.

For further reading, consider exploring more complex probability scenarios and using tools such as Permutations and Combinations to solve similar problems.