A bag contains five white balls and three black balls. Your goal is to draw two white balls.
(a) You simultaneously draw two balls at random. What is the probability that they are both white?
(b) You simultaneously draw two balls at random. Once you have drawn two balls, you put back any black balls, and redraw so that you again have two drawn balls. What is the probability that you now have two white balls? (Include the probability that you chose two white balls on the first draw.)
(a) The probability that you draw two white balls simultaneously is 5/8 x 4/7 = 10/56.
(b) The probability that you draw two white balls on the first draw and then redraw two white balls is 5/8 x 4/7 x 5/8 = 25/224.
To find the probability of drawing two white balls, we can use the concept of combinations.
(a) In this case, you want to draw two white balls out of a total of eight (five white balls and three black balls). The total number of ways to draw two balls from eight is given by the combination formula: C(8,2) = 8! / (2!(8-2)!) = 28.
Now, we need to find the number of ways to draw two white balls. Since there are five white balls in the bag, the number of ways to choose two white balls is given by the combination formula again: C(5,2) = 5! / (2!(5-2)!) = 10.
Therefore, the probability of drawing two white balls simultaneously is 10/28 = 5/14.
(b) In this scenario, after drawing two balls, any black ball that was drawn is put back into the bag, and you redraw two balls. This means that for the second draw, the number of white and black balls remains the same as in the first draw.
The probability of drawing two white balls in the first draw is 5/8 (since there are five white balls and a total of eight balls).
Now, let's consider the probabilities for the second draw. If you drew two white balls in the first draw, the probability of drawing two white balls in the second draw is still the same as in the first draw, which is 5/8.
If you drew one white ball and one black ball in the first draw, putting back the black ball means you still have seven balls in the bag, with four white balls and three black balls. So, the probability of drawing two white balls in the second draw is given by the combination formula: C(4,2) / C(7,2) = 6 / 21 = 2/7.
Finally, we need to consider the case where both balls drawn in the first attempt were black. In this case, you put both black balls back into the bag, so the number of balls for the second draw is the same as the original scenario (five white balls and three black balls). Therefore, the probability of drawing two white balls in the second draw is 5/8.
Now, we need to consider the probabilities of each scenario and their respective probabilities of occurring:
1. Drawing two white balls in the first draw: Probability = 5/8
Probability of drawing two white balls in the second draw = 5/8
2. Drawing one white ball and one black ball in the first draw: Probability = (5/8) * (3/8)
Probability of drawing two white balls in the second draw = 2/7
3. Drawing two black balls in the first draw: Probability = (3/8) * (2/8)
Probability of drawing two white balls in the second draw = 5/8
Now, we can calculate the overall probability by multiplying the probability of each scenario by its respective probability of occurring and summing them up:
Overall probability = (5/8) * (5/8) + (5/8) * (3/8) * (2/7) + (3/8) * (2/8) * (5/8)
Simplifying this expression will give us the final answer.
(a) To find the probability that both balls drawn are white, we need to consider the total number of possible outcomes and the number of favorable outcomes.
1) Total number of possible outcomes:
Since there are 5 white balls and 3 black balls, the total number of balls in the bag is 8. When drawing 2 balls at random, there are a total of C(8, 2) = 28 possible outcomes.
2) Number of favorable outcomes:
To draw 2 white balls, we need to choose 2 white balls out of the 5 available in the bag. The number of ways to choose 2 white balls out of 5 is C(5, 2) = 10.
Therefore, the probability that both balls drawn are white is:
P(both white) = favorable outcomes / total outcomes = 10 / 28 = 5/14.
(b) Similarly to part (a), we will consider the number of possible outcomes and the number of favorable outcomes.
1) Total number of possible outcomes:
After drawing 2 balls and putting back any black balls, the total number of balls in the bag remains the same - 8. So, the total number of possible outcomes is still C(8, 2) = 28.
2) Number of favorable outcomes:
To have two white balls after redrawing, we have two possible scenarios:
- Scenario 1: We drew 2 white balls on the first draw, and both white balls were also drawn on the second draw.
- Scenario 2: We drew 1 white ball and 1 black ball on the first draw, and both white balls were drawn on the second draw.
- Scenario 1:
The probability of drawing 2 white balls on the first draw is the same as in part (a): P(both white) = 5/14.
Once the two white balls are put back, the total number of balls in the bag remains 8 with 5 white and 3 black balls. Thus, the probability of drawing 2 white balls on the second draw is also P(both white) = 5/14.
Therefore, the probability of Scenario 1 is P(Scenario 1) = P(both white) * P(both white) = (5/14) * (5/14) = 25/196.
- Scenario 2:
The probability of drawing 1 white and 1 black ball on the first draw is given by: P(1 white, 1 black) = (C(5, 1) * C(3, 1)) / C(8, 2) = (5 * 3) / 28 = 15/28.
Once we put back the black ball and redraw, the total number of balls in the bag remains 8, with 6 white (including the one previously drawn) and 2 black balls. Therefore, the probability of drawing 2 white balls on the second draw is P(both white) = 6/8 = 3/4.
Thus, the probability of Scenario 2 is P(Scenario 2) = P(1 white, 1 black) * P(both white) = (15/28) * (3/4) = 45/112.
Overall, the probability of having two white balls after redrawing is the sum of the probabilities of the two scenarios:
P(both white after redrawing) = P(Scenario 1) + P(Scenario 2) = 25/196 + 45/112 = 355/784.