# A box contains 10 red markers, 4 green markers, 2 yellow markers, and 4 blue markers.?

(a) With replacement, what are the chances of getting 4 red markers in a row?

(b) With replacement, what are the chances of getting 1 red marker, then a blue marker, then a yellow marker, and finally, a green marker?

(c) Without replacement, what are the chances of getting 1 red marker, then a blue marker, then a yellow marker, and finally, a green marker?

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This looks like it ought to have been posted in the Mathematics section.

(a) With replacement, what are the chances of getting 4 red markers in a row?

"With replacement" means we put back each marker before randomly pulling out the next one. Half the markers (10 out of 20) are red, so the probably of getting 4 red ones in a row is

(1/2) (1/2) (1/2) (1/2) = 1/16

(b) With replacement, what are the chances of getting 1 red marker, then a blue marker, then a yellow marker, and finally, a green marker?

This is similar to part (a), except that the probabilities of different colors vary:

10/20 = 1/2 for red,

4/20 = 1/5 for green and blue, and

2/20 = 1/10 for yellow.

Because we're drawing WITH replacement, the probability of completing this sequence, in order, is

(1/2) (1/5) (1/10) (1/5) = 1/500

(c) Without replacement, what are the chances of getting 1 red marker, then a blue marker, then a yellow marker, and finally, a green marker?

WITHOUT replacement, we're drawing from a smaller set of markers at each step:

(1/2) (4/19) (2/18) (4/17)

= (4 * 2 * 4) / (2 * 19 * 18 * 17)

= 8 / (19 * 9 * 17)

= 8/2,907

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