A box contains exactly five chips, three red and
two white. Chips are randomly
• removed one at a time without replacement until all the
red chips are drawn or
• all the white chips are drawn. What is the probability that the last chip drawn
is white?
• (
2/5) x (1/4) = 1/10
The probability that the last chip drawn is white can be found by considering two cases:
1) The first four chips drawn are red
2) One or more red chips are drawn before the last white chip is drawn.
Let's calculate the probability for each case:
1) In this case, we need to consider the probability of drawing the four red chips first, followed by the last white chip.
The probability of drawing the first red chip is 3/5. The probability of drawing a second red chip, given that the first chip was red, is 2/4 (since there are now 4 chips remaining and 2 are red). Continuing in the same manner, the probability of drawing the four red chips in a row is:
(3/5) * (2/4) * (1/3) * (0/2) = 0
Since there are no chips left to draw after the four red chips have been drawn, the probability of the last chip being white in this case is 0.
2) In this case, we can have different scenarios where one or more red chips are drawn before the last white chip is drawn.
Scenario 1: The first three chips drawn are red, and the fourth chip is white.
The probability of drawing the first red chip is 3/5. The probability of drawing a second red chip, given that the first chip was red, is 2/4. The probability of drawing a third red chip, given that the first two chips were red, is 1/3. Finally, the probability of drawing a white chip as the fourth chip is 2/2, since there are only white chips left.
(3/5) * (2/4) * (1/3) * (2/2) = 1/10
Scenario 2: The first two chips drawn are red, the third chip is white, and the fourth chip drawn is red.
Following the same logic as above, the probability of this scenario is also 1/10.
Scenario 3: The first chip drawn is red, the second chip is white, and the third and fourth chips are red.
The probability of drawing the first red chip is 3/5. The probability of drawing a white chip as the second chip is 2/4. The probability of drawing a red chip as the third and fourth chips is 2/3 * 1/2 = 1/3.
(3/5) * (2/4) * (1/3) * (1/2) = 1/20
Adding the probabilities of all three scenarios:
1/10 + 1/10 + 1/20 = 3/20
Therefore, the probability that the last chip drawn is white is 3/20.
To find the probability that the last chip drawn is white, we need to consider two possible scenarios:
Scenario 1: All red chips are drawn before any white chips are drawn.
Scenario 2: All white chips are drawn before all red chips are drawn.
Let's calculate the probability of each scenario and add them together to get the final probability.
Scenario 1:
To calculate the probability of all red chips being drawn first, we need to find the probability of drawing a red chip at each step. Since there are 3 red chips out of 5 total chips, the probability of drawing a red chip on the first draw is 3/5.
After the first red chip is drawn, there are 4 chips remaining in the box, with 2 of them being red chips. So, the probability of drawing a red chip on the second draw, given that the first chip was red, is 2/4.
Similarly, after the second red chip is drawn, there are 3 chips remaining in the box, with 1 of them being red. So, the probability of drawing a red chip on the third draw, given that the first two chips were red, is 1/3.
Since we need to draw all red chips before any white chips, we multiply the probabilities of each step:
(3/5) * (2/4) * (1/3) = 6/60 = 1/10.
Scenario 2:
To calculate the probability of all white chips being drawn first, we follow the same approach as in scenario 1.
The probability of drawing a white chip on the first draw is 2/5.
After the first white chip is drawn, there are 4 chips remaining in the box, with 1 of them being white. So the probability of drawing a white chip on the second draw, given that the first chip was white, is 1/4.
Similarly, the probability of drawing a white chip on the third and final draw, given that the first two chips were white, is 1/3.
So, the probability of drawing all white chips before all red chips is:
(2/5) * (1/4) * (1/3) = 2/60 = 1/30.
To find the probability that the last chip drawn is white, we add together the probabilities of both scenarios:
1/10 + 1/30 = 3/30 + 1/30 = 4/30 = 2/15.
Therefore, the probability that the last chip drawn is white is 2/15.