You buy a pack of 100 cards. due to manufacturing inaccuracies, the pack may actually contain 99, 100 or 101 cards (with each option being equally likely). if you purchase 3 packs, what is the probability that you have exactly 300 cards?

Let's think of the difference from the desired value: if we get a 99 cards pack we'll count it as a "-1", a 100 cards pack counts as a "0", and a 101 cards pack counts as "+1".

We buy 3 packs, so there are 27 possible scenarios (each of the three packs can either be a "-1", "0" or "+1".

Let's count how many of these sum to zero: these are the cases where we have 300 card in total.

These cases are:

Just to make things clear, the first option means that you buy three 100 cards pack, the second means that the first pack is a 100 cards one, the second is a 101 cards one, and the third is a 99 one, and so on.

The pattern is always the same: there has to be a 100 cards pack, and then you have a 99 cards one and a 101 cards one, so that they balance, plus the "special case" when you buy three "correct" packs with 100 cards each.

So, there are 7 cases when you have 300 cards, and we already said that there are 27 possible scenarios. So, the probability of having 300 cards is 7/20.

p> q if you calculate p is100 q is 79

can you add an image ?

step-by-step explanation: