Respuesta :

Answer:

The probability of drawing a red block, keeping it, and drawing another is approximately 2/7. The probability of draawing a blue bock, keeping it, and drawing another blue block is approximately 1/7. (The question is worded very poorly. I'm assuming that in the second scenario, all blocks are still in the bag.)

Step-by-step explanation:

Answer:

We have 4 red blocks and 3 blue blocks:

Then, if we draw one of those at random, the probability of each color is the number of blocks of the given color divided the total number of blocks.

Then the probability of drawing a red block is

p1 = 4/7

Now, the probability of drawing another red block is obtained in the same way; but we now take away one red block. so the number of red blocks is 3, and the total number of blocks is 6, so the probability is:

p2 = 3/6 = 1/2

Now, the probability of both events happening is equal to the product of these probabilities; so P = p1*p2 = 4/7*1/2 = 4/14 = 2/7

b)

The probability of first drawing a blue block is

p1 = 3/7

and now we have 2 blue blocks and 6 red blocks, now the probability of drawing a red block is:

p2 = 4/6 = 2/3

The joint probability is P = 3/7*2/3 = 2/7

So you can see that the probability in a) and b) are the equal.

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