Joe has a bag containing 8 red sweets, 9 yellow ones and 11 green. He takes out a sweet and eats it, he takes out a second sweet. What is the probability that both the sweets are red?

2/27

Step-by-step explanation:

This is a probability question and we are asked to estimate a particular probability. We proceed as follows:

The total number of sweets is given as 8 + 9+ 11= 28 sweets

He takes out a sweet to eat, the probability of this being a red sweet would be P(r1) = 8/28 = 2/7

Now he takes another sweet, we are asked to calculate the probability that this sweet is also red. Now after taking the first sweet, the number of sweets is now 27, while the number of red sweets is now 7. Hence the probability of having a red sweet taken in the second case would be p(r2) = 7/27

Now, the probability of both being red sweets can be calculated by multiplying both = 7/27 * 2/7 = 2/27

"image"

step-by-step explanation:

in most of these problems, the notation is that the point with a "prime" is the image of the point without. that is, b' is the image of b.

here, there is no dilation or reflection. there is simply translation (with no direction shown). which figure you call the "preimage" and which one you call the "image" is up to you. as a matter of convention, the naming is as described above.

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