6 black counters and 4 white counters in bag A.
7 black counters and 3 white counters in bag B.
5 black counters and 5 white counters in bag C.
Tarek takes at random a counter from bag A and puts the counter in bag B. He then takes at random a counter from bag B and put the counter in bag C.
What is the probability that there are now more black counters than white counters in bag C.
Thank you very very much.
I’m a bit confused as to where the 24/110 came from?
Thank you for explaining it so clearly. x
yeah i agree w/ tom where does 24/110 come from do you mean 28/110??
Well, we're going to do some counter-crazy calculations here!
Let's start by figuring out the probabilities of Tarek picking a black or white counter from bag A.
The probability of picking a black counter from bag A is 6/10, since there are 6 black counters out of a total of 10.
The probability of picking a white counter from bag A is 4/10, since there are 4 white counters out of a total of 10.
Now, let's look at the probabilities after Tarek puts a counter in bag B.
If Tarek picked a black counter from bag A, the probabilities in bag B would be 7 black counters out of 11 total counters.
If Tarek picked a white counter from bag A, the probabilities in bag B would be 3 white counters out of 11 total counters.
Now, let's move on to bag C after Tarek puts a counter from bag B.
If Tarek picked a black counter from bag B, the probabilities in bag C would be 6 black counters out of 11 total counters.
If Tarek picked a white counter from bag B, the probabilities in bag C would be 5 white counters out of 11 total counters.
Now, we can calculate the probability of there being more black counters than white counters in bag C.
For there to be more black counters than white counters, either there can be 6 or more black counters and 5 or less white counters, or there can be 7 or more black counters and 4 or less white counters.
So, we need to calculate the probability of having 6 or more black counters and 5 or less white counters, plus the probability of having 7 or more black counters and 4 or less white counters.
The probability of having 6 or more black counters and 5 or less white counters is the probability of having 6 black counters (6/11) plus the probability of having 7 black counters (1/11) multiplied by the probability of having 4 or less white counters (5/11).
The probability of having 7 or more black counters and 4 or less white counters is the probability of having 7 black counters (1/11) multiplied by the probability of having 3 or less white counters (4/11) plus the probability of having 8 black counters (0, since there are only 7 possible black counters in bag B) multiplied by the probability of having 4 or less white counters (5/11).
I hope you brought a calculator, because now we need to do some math! Add the two probabilities together and you'll have your answer!
(Drumroll please...)
Gosh, I still can't believe I made you go through all of that. Okay, let's break it down into simplified fractions.
The probability that there are now more black counters than white counters in bag C is:
(6/11) + ((1/11) * (5/11)) + ((1/11) * (4/11)) = (6/11) + (5/121) + (4/121) = 71/121
So, there you have it. The probability is 71/121.
I hope I didn't confuse you too much with all those counters flying around. Good luck with your probability calculations!
To find the probability that there are more black counters than white counters in bag C after Tarek's actions, we need to consider all possible outcomes and determine how many of them satisfy the condition.
First, let's calculate the total number of possible outcomes. The number of possible outcomes corresponds to the number of ways Tarek can take counters from each bag. We start with bag A, which has 6 black counters and 4 white counters. The probability of choosing a black counter from bag A is 6/10, and the probability of choosing a white counter is 4/10.
Next, we move to bag B, which now contains one counter taken from bag A. After Tarek adds a counter from bag A to bag B, there will be a total of 11 counters in bag B (since it originally had 7 black counters and 3 white counters). Therefore, the probability of choosing a black counter from bag B is now 7/11, and the probability of choosing a white counter is 4/11.
Finally, we consider bag C, which initially has 5 black counters and 5 white counters. After Tarek adds a counter from bag B to bag C, bag C will have a total of 11 counters (since it originally had 5 of each color). Thus, the probability of choosing a black counter from bag C is 5/11, and the probability of choosing a white counter is also 5/11.
To determine the probability that there are more black counters than white counters in bag C, we need to calculate the probabilities of the different possibilities. These include scenarios where there are more blacks than whites, equal numbers of black and white counters, and more whites than blacks in bag C.
Let's consider the possible scenarios:
1. More black counters in bag C: This corresponds to having 6 or more black counters (since initially there are 5 in bag C) and 5 or fewer white counters (since initially there are 5 in bag C). To calculate this probability, we need to consider the different combinations of black and white counters that satisfy this condition.
- There are 6 ways to have 6 black counters and 0 white counters (BBBBBB).
- There are 5 ways to have 7 black counters and 0 white counters (BBBBBBB).
- There are 4 ways to have 7 black counters and 1 white counter (BBBBBBWW).
- There are 3 ways to have 8 black counters and 0 white counters (BBBBBBBB).
- There are 2 ways to have 8 black counters and 1 white counter (BBBBBBBWW).
- There is 1 way to have 9 black counters and 0 white counters (BBBBBBBBB).
- There is 1 way to have 9 black counters and 1 white counter (BBBBBBBBWW).
- There is 1 way to have 10 black counters and 0 white counters (BBBBBBBBBB).
- There is 1 way to have 10 black counters and 1 white counter (BBBBBBBBBWW).
- There is 1 way to have 10 black counters and 2 white counters (BBBBBBBBBWW).
So, the probability of having more black counters in bag C is (6 + 5 + 4 + 3 + 2 + 1 + 1 + 1 + 1 + 1) / (10 * 11 * 11) = 0.3636 or 36.36%.
2. Equal number of black and white counters in bag C:
- There are 2 ways to have 5 black counters and 5 white counters (BBBBBWWWWW and BBBBWWWWWW).
So, the probability of having an equal number of black and white counters in bag C is 2 / (10 * 11 * 11) = 0.0182 or 1.82%.
3. More white counters in bag C:
- There is 1 way to have 5 black counters and 6 white counters (BBBBBWWWWW and BBBBWWWWWWW).
- There is 1 way to have 5 black counters and 7 white counters (BBBBBWWWWWW and BBBBWWWWWWWW).
So, the probability of having more white counters in bag C is 2 / (10 * 11 * 11) = 0.0182 or 1.82%.
Adding up these probabilities, we find that the probability of having more black counters than white counters in bag C is 0.3636 + 0.0182 = 0.3818 or 38.18%.
Therefore, the probability that there are more black counters than white counters in bag C after Tarek's actions is 38.18%.
3/5
work out the probability of two scenarios
1. the probability of picking a black counter from bag a then a black counter from bag B
6/10 * 8/11 = 48/110
2 The probability of picking a white counter form bag A then a black counter from bag B
4/10 * 7/11 = 28/110
In both scenarios you end with a black counter to put in bag C making bad C have more black counters than white. you combine the answers and get the probability of having more black counters in Bag C
24/110 + 48/110 = 72/110