An urn contains 12 balls, five of which are red. Selection of a red ball is desired and is therefore considered to be a success. If three balls are selected, what is the expected value of the distribution of the number of selected red balls

Assume there are equal amount of a lot of red and other color balls in the urn for you to select r=red and o=other, all the possible outcome for selecting 3 in sequence is 1r2r3r, 1r2r3o,1r2o3r, 1o2r3r, 1r2o3o, 1o2r3o, 1o2o3r, 1o2o3o. In terms of expected value, 1 case for all 3 reds, 3 cases for 2 reds, 3 cases for 1 red, and 1 case for 0 red. Equal large amount of red and other means all 8 outcomes has equal possibilities so each is 1/8 and 3 reds will have possibility of 1/8, 2 reds will have 3/8, 1 red will have 3/8 and no red will have 1/8. However the question tells you the red and other don’t have equal amounts, 5/12 possibility for each pick for red and 7/12 possibility for other. So three reds becomes 5/12*5/12*5/12; 2 reds becomes 5/12*5/12*7/12 multiply by 3 (since 3 cases); 1 red is 5/12*7/12*7/12 multiply by 3 (since 3 cases); 0 red is 7/12*7/12*7/12. Out of every 10 picks, you will expect to pick all 3 reds with the chance less than once, 2 reds with the chance more than 3 times and 1 red with the chance more than 4 times and no red with the chance less than 2 times. (I forgot how to exactly represent expected value) and possibility distribution is estimated 3 reds around 7%, 2 reds around 30%, 1 red around 42% and no red around 20%.

Just a question- could you organize the data better? i just cannot tell which number are for which bakery. i will be happy to answer this question if you could do this for me! : )