There will still be 34.5 L of water in the two pails combined because no water is lost.
Thus the total remains 34.5 L after transferring water between them.
Let the final amount in the smaller pail be x.
Then the larger pail contains 9x.
So x + 9x = 34.5.
That simplifies to 10x = 34.5.
Dividing both sides by 10 gives x = 3.45.
Therefore the smaller pail held 3.45 L at the end.
Because 0.68 L was poured into it, its initial volume was 2.77 L (3.45 minus 0.68)
We can also deduce the larger pail initially contained 31.73 L
(either 35 minus 2.77, or 9 times 3.45 plus 0.68)
Response:
Step-by-step explanation:
Assuming there are 100 sour candies, thus-
since 26% of the candies are grape, it follows that 26% of 100 candies means we have 26 grape candies
Consequently, the remaining candies that are not grape amount to 100-26 = 74
Applying the multiplication principle:
P(A ∩ B) = P(A)/ P(B|A)
Initially, there are 26 grape candies, therefore the probability of selecting the first grape candy = 26C1 = 26
After choosing the first one, we put the selected grape candy back, so there are still 100 candies total P(B|A) = 100C3 = 100 x 99 x 98 x 97!/3! X 97!
= 50 x 33 x 98
Thus, the probability becomes 1/ 50 x 33 x 98 x 26
= 1/4204200
Answer:
The response is "I made an error."
Step-by-step explanation:
It's not possible to have a negative probability; it must always be zero or greater.
There are several possible outcomes. The initial composition of the urns is as follows: Urn 1 contains 2 red chips and 4 white chips, totaling 6 chips, whereas Urn 2 has 3 red and 1 white, amounting to 4 chips. When a chip is drawn from the first urn, the probabilities are as follows: for a red chip, it is probability is (2 red from 6 chips = 2/6 = 1/2); for a white chip, it is (4 white from 6 chips = 4/6 = 2/3). After the chip is transferred to the second urn, two scenarios can arise: if the chip drawn from the first urn is white, then Urn 2 will contain 3 red and 2 white chips, making a total of 5 chips, creating a 40% chance for drawing a white chip. Conversely, if a red chip is drawn first, Urn 2 will contain 4 red and 1 white chip, which results in a 20% chance of drawing a white chip. This scenario exemplifies a dependent event, as the outcome hinges on the type of chip drawn first from Urn 1. For the first scenario, the combined probability is (the probability of drawing a white chip from Urn 1) multiplied by (the probability of drawing a white chip from Urn 2), equaling 26.66%. For the second scenario, the probabilities yield a value of 6%.
