A bag contains 8 white and 3 black counters. Two counters are drawn, one after the other. Find the probability of drawing one white and one black counter, in any order. i) If the first counter is replaced ii) if the first counter is not replaced.
White = 8/11
Black = 3/11
1) If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.
P(BW) = 3/11 * 8/11
P(WB) = 8/11 * 3/11
Either-or probabilities are found by adding the individual probabilities.
2) P(BW) = 3/11 * 8/(11-1)
P(WB) = 8/11 * 3/(11-1)
Well, isn't this an exciting game of counter-drawing! Let's calculate the probabilities and have some fun, shall we?
i) If the first counter is replaced:
When the first counter is drawn, there are 8 white counters out of a total of 11 counters, so the probability of drawing a white counter on the first attempt is 8/11.
After the first counter is put back, the total number of counters remains the same (11), but the number of white counters decreases to 7, and the number of black counters stays the same (3).
So, when drawing the second counter, the probability of drawing a black counter is now 3/11.
Since we're looking for the probability of drawing one white and one black counter in any order, we can multiply the probabilities of drawing a white and a black counter separately and then double the result (as the order could be black-white or white-black).
So, the probability is (8/11) * (3/11) * 2 = 48/121.
ii) If the first counter is not replaced:
When the first counter is drawn, there are 8 white counters out of a total of 11 counters, so the probability of drawing a white counter on the first attempt is 8/11.
After the first counter is not replaced, the total number of counters decreases to 10, and the number of white counters also decreases to 7, while the number of black counters stays the same (3).
So, when drawing the second counter, the probability of drawing a black counter is now 3/10.
Again, since we're looking for the probability regardless of the order, we can multiply the probabilities and double the result: (8/11) * (3/10) * 2 = 48/110.
And there you have it, the probabilities of drawing one white and one black counter, in any order, with or without replacing the first counter. Now you can impress your friends with your incredible counter-drawing knowledge!
i) If the first counter is replaced:
To find the probability of drawing one white and one black counter in any order when the first counter is replaced, we can consider two possibilities:
1. Drawing a white counter followed by a black counter.
2. Drawing a black counter followed by a white counter.
Given that there are 8 white counters and 3 black counters in the bag, the probability of drawing a white counter on the first draw is 8/11, and the probability of drawing a black counter on the second draw is also 3/11 (since the first counter is replaced).
Therefore, the probability of drawing one white and one black counter in any order when the first counter is replaced is:
P(white then black) + P(black then white)
= (8/11) * (3/11) + (3/11) * (8/11)
= (24/121) + (24/121)
= 48/121
ii) If the first counter is not replaced:
To find the probability of drawing one white and one black counter in any order when the first counter is not replaced, we can follow a similar approach as in case i), but the probabilities will change slightly.
For the first draw, the probability of drawing a white counter is 8/11, and for the second draw, the probability of drawing a black counter is 3/10 (since the first counter is not replaced and there is one less counter in total).
Therefore, the probability of drawing one white and one black counter in any order when the first counter is not replaced is:
P(white then black) + P(black then white)
= (8/11) * (3/10) + (3/11) * (8/10)
= (24/110) + (24/110)
= 48/110
= 24/55
To find the probability in both cases, we need to calculate the probability of drawing a white counter first and then a black counter, and also the probability of drawing a black counter first and then a white counter.
Let's start with the case where the first counter is replaced:
i) If the first counter is replaced:
The probability of drawing a white counter first and then a black counter can be calculated as follows:
P(White first, Black second) = P(White) * P(Black) = (8/11) * (3/11) = 24/121
Similarly, the probability of drawing a black counter first and then a white counter can be calculated as follows:
P(Black first, White second) = P(Black) * P(White) = (3/11) * (8/11) = 24/121
To find the probability of drawing one white and one black counter in any order, we need to add the probabilities of both cases:
P(One white and one black, with replacement) = P(White first, Black second) + P(Black first, White second) = 24/121 + 24/121 = 48/121
Now, let's consider the case where the first counter is not replaced:
ii) If the first counter is not replaced:
The probability of drawing a white counter first and then a black counter can be calculated as follows:
P(White first, Black second) = P(White) * P(Black) = (8/11) * (3/10) = 24/110
Similarly, the probability of drawing a black counter first and then a white counter can be calculated as follows:
P(Black first, White second) = P(Black) * P(White) = (3/11) * (8/10) = 24/110
To find the probability of drawing one white and one black counter in any order, we need to add the probabilities of both cases:
P(One white and one black, without replacement) = P(White first, Black second) + P(Black first, White second) = 24/110 + 24/110 = 48/110 = 24/55
Therefore, the probability of drawing one white and one black counter, in any order, is:
i) If the first counter is replaced: 48/121
ii) If the first counter is not replaced: 24/55