Math Probability Question.?

5 candy canes,

3 of which are cracked.

2 not cracked

(i love candy canes)

2/5 = the probability of selecting 1 of the 2 no-crack canes on the 1st selection

now only 4 canes remain

you need to select the other no-crack cane

only 1 remains and only 4 canes remain

1/4

so for both events to occur = 2/5 * 1/4 = 2/20 = 1/10 or 0.10

**********

i would rather solve this using Pascal's Triangle (a combination method)

as this is a hypergeometric distribution (really - the probability of selecting a no-crack cane changes when there is no replacement)

(2 no-crack choose 2) * (3 cracked choose 0) / 5 canes choose 2

2C2 * 3C0 / 5C2 =

1 * 1 / 10 = 1/10

values easily seen in the triangle!

in agreement with the other method

*****

when parameters are way different the combination method is super-easy and fun

where the think-it-thru-method becomes bogged down with lots of math and ways something can happen, imo