We have four boxes, each of which contains ten coloured balls:

Box 1 contains 4 red balls, 5 green balls and 1 yellow ball

Box 2 contains 3 red balls, 5 green balls and 2 yellow balls

Box 3 contains 2 red balls, 5 green balls and 3 yellow balls

Box 4 contains 1 red ball, 5 green balls and 4 yellow balls

If we randomly select one ball from each box, what is the probability of selcecting exactly one red ball?


If we randomly select one ball from each box, what is the probability of selcecting at one red ball?

First, we have to realize that it's asking for EXACTLY 1 red one. This means that if you pick a red one from the first box, the rest of the boxes CAN'T be red. The next thing we have to realize is that there are multiple ways of obtaining a red ball (i.e. we can get it from the first box OR the second box). This means, we have to add our probabilities.

So, let's calculate the probability that we pick a red ball from the first box. That's 0.4. Then, since the rest can't be red, we have to calculate the probability of NOT picking a red ball for the rest. That's 0.7 for the second box, .8 for the third box, and .9 for the 4th box.

So, the probability of picking one red marble from the first box and the rest of the boxes non-red is: 0.4*.7*.8*.9

As I said previously, there are multiple way so of picking a red ball, so you have to do a similar exercise with the other 3 possibilities. I'll let you contemplate everything so that you get the hang of these types of problems.

For the last one, I'll give you this hint.
Probability of AT LEAST 1 = Probability of EXACTLY 1+Probability of EXACTLY 2+Probability of EXACTLY 3+Probability of EXACTLY 4

Also, remember that Probability of EXACTLY 1+Probability of EXACTLY 2+Probability of EXACTLY 3+Probability of EXACTLY 4+Probability of EXACTLY 0 is equal to 1.

Let me know if you need anymore help.