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 ?
number of ways that the 3 reds and the 2 whites
can be arranged = 5!/(3!2!) = 10
Now place a white at the end, then the number of ways to arrange the remaining 4 is 4!/(2!2!) = 6
so prob that the last one will be white = 6/10 = 3/5
To find the probability that the last chip drawn is white, we need to consider the different possible outcomes.
There are two scenarios in which the last chip drawn is white:
1. Scenario 1: The two white chips are drawn before the three red chips.
2. Scenario 2: The three red chips are drawn before the two white chips.
Let's calculate the probabilities of these two scenarios separately:
1. Scenario 1:
In this case, we need to find the probability of drawing two white chips before three red chips. The probability of drawing the first white chip is 2/5 (since there are two white chips out of five total chips). After the first white chip is drawn, there is one white chip remaining out of the four remaining chips. Therefore, the probability of drawing a second white chip is 1/4. Multiplying these probabilities together gives us (2/5) * (1/4) = 1/10.
2. Scenario 2:
Now we need to find the probability of drawing three red chips before the two white chips. The probability of drawing the first red chip is 3/5 (since there are three red chips out of five total chips). After the first red chip is drawn, there are two red chips remaining out of the four remaining chips. Therefore, the probability of drawing a second red chip is 2/4. Similarly, the probability of drawing the third red chip is 1/3. Multiplying these probabilities gives us (3/5) * (2/4) * (1/3) = 1/10.
Since these two scenarios are mutually exclusive (i.e., they cannot happen at the same time), we can add the probabilities of these two scenarios to find the total probability:
1/10 + 1/10 = 2/10 = 1/5
Therefore, the probability that the last chip drawn is white is 1/5.
To find the probability that the last chip drawn is white, we need to consider the different scenarios in which this can happen.
Scenario 1: The last chip drawn is the first white chip.
In this scenario, the sequence of draws could be:
1. Red, Red, White.
2. Red, White.
Scenario 2: The last chip drawn is the second white chip.
In this scenario, the sequence of draws could be:
1. White, Red, Red.
2. White, White.
In both scenarios, the last chip drawn is white. Now, we need to calculate the probability of each scenario occurring and then add them together to get the overall probability.
First, let's calculate the probability for Scenario 1:
The probability of drawing the first red chip is 3/5 (as there are 3 red out of total 5 chips).
The probability of drawing the second red chip (without drawing a white chip in between) is 2/4 (as there are 2 red chips left out of 4 remaining chips).
The probability of drawing the white chip is 2/3 (as there are 2 white chips left out of 3 remaining chips).
Therefore, the probability for Scenario 1 is (3/5) * (2/4) * (2/3) = 12/60 = 1/5.
Now, let's calculate the probability for Scenario 2:
The probability of drawing the first white chip is 2/5 (as there are 2 white out of total 5 chips).
The probability of drawing the second white chip (without drawing a red chip in between) is 1/4 (as there is 1 white chip left out of 4 remaining chips).
Therefore, the probability for Scenario 2 is (2/5) * (1/4) = 2/20 = 1/10.
Finally, we add the probabilities of both scenarios to get the overall probability that the last chip drawn is white:
1/5 + 1/10 = 3/10.
Hence, the probability that the last chip drawn is white is 3/10.