1) Our marbles will be blue, red, and green. You need two fractions that can be multiplied together to make 1/6. There are two sets of numbers that can be multiplied to make 6: 1 and 6, and 2 and 3. If you give the marbles a 1/1 chance of being picked, then there's no way that a 1/6 chance can be present So we need to use a 1/3 and a 1/2 chance. 2 isn't a factor of 6, but 3 is. So we need the 1/3 chance to become apparent first. Therefore, 3 of the marbles will need to be one colour, to make a 1/3 chance of picking them out of the 9. So let's say 3 of the marbles are green. So now you have 8 marbles left, and you need a 1/2 chance of picking another colour. 8/2 = 4, so 4 of the marbles must be another colour, to make a 1/2 chance of picking them. So let's say 4 of the marbles are blue. We know 3 are green and 4 are blue, 3 + 4 is 7, so the last 2 must be red.
The problem could look like this:
A bag contains 4 blue marbles, 2 red marbles, and 3 green marbles. What are the chances she will pick 1 blue and 1 green marble?
You should note that picking the blue first, then the green, will make no difference to the overall probability, it's still 1/6. Don't worry, I checked
2) a - 2% as a probability is 2/100, or 1/50. The chance of two pudding cups, as the two aren't related, both being defective in the same packet are therefore 1/50 * 1/50, or 1/2500.
b - 1,000,000/2500 = 400
400 packages are defective each year
Answer:
A 28 because 7 times four
Step-by-step explanation:
If A is a subset of B, then all elements from a set A are in a set B.
The number 8 from {3, 8, 6} is not in the set {1, 2, 3, 4, 5, 6}. Therefore {3, 8, 6} is not a subset of {1, 2, 3, 4, 5, 6}.
5x + 2y = 7 . . . (1)
y = x + 1 . . . (2)
Putting (2) into (1) gives
5x + 2(x + 1) = 7 => 5x + 2x + 2 = 7 => 7x = 7 - 2 = 5 => x = 5/7
From (2) y = 5/7 + 1 = 12/7
Therefore, solution is {(5/7, 12/7)}
I would say a. is not very common.
I did not calculate the rest, I am just canceling out one for you
