Question

there are ten marbles in a bag. 3 are green and 7 are red. you mix them up and pick one and note the color then REPLACE it back in the bag, now you select a second marble and note its color. what is the probability you select 2 red marbles? 2 green marbles or one of each color?

Answer 1

Probability you pick one green marble with one trial is 3/10 because 3 of the 10 marbles are green. Similarly, for red, it will be 7/10.

These are independent trials, so we can multiply the probabilities.

To get two red marbles in a row we have \[\frac{7}{10}*\frac{7}{10} = 0.49\] or 49%.

Two green marbles in a row would be \[\frac{3}{10}*\frac{3}{10}=0.09\] or 9%

The final question is a bit trickier, because there are two ways to pick one of each color: red then green, and green then red. We compute each case separately, then add them to get the probability of either one happening.

Red then green: \[\frac{7}{10}*\frac{3}{10}=0.21\]
Green then red: \[\frac{3}{10}*\frac{7}{10}=0.21\]

Adding the two together we get 0.42 or 42%. We can check that this result makes sense by noticing that the cases we didn't count (red then red, and green then green) added to the cases we did count make up the set of all possible cases, and the probabilities add up to 1 or 100%. If we add 0.42 + 0.49 + 0.09 (the latter two being the probabilities we found earlier), it adds up to 1, so we haven't under or over-counted anything.