Imagine that we have a single muon. It has a life span of 2.2 microseconds on average. So normally, after that amount of time, the muon will decay. Now imagine we have two groups of muons. Group A has 20 muons and the other group, B , 10 muons. Since muons have a half life of 2.2 microseconds, group A will lose 10 muons in that time but group B will lose 5 muons. Why is this so? If the a sinle muons decays in 2.2 microseconds, normally, all the muons should decay in that time as they are all muons and so decay in 2.2 microseconds. Group A lost more muons in the same time than group B. This can only be if the muons in group a decayed faster than the muons in group B. how can this be? Is the life of muons inversely proportional to the amount of muons present?
Imagine two groups of students, where group A has 20 members and group B has 10. Each student flips a coin, and we find that 10 coins from group A and 5 coins from group B land face-up.
Does the probability of the coin landing face-up depend on the size of the group? No, of course not. The probability of a coin landing face-up is simply 1/2, which means that we'd expect to see it in about half of the coins in a group of any size. Larger groups flip more heads because more coins are being tossed, that's all - and precisely the same is true for radioactive decay.
No, muon decay is a statistical process with a constant half life. After 2.2 microseconds there is a 50% chance that the muon has decayed. This means that on average half of the muons have decayed. For group A half is 10 and for B it is 5. Note that muons that are generated in the atmosphere by cosmic rays live much longer as they travel at near light speed.
Imagine you had two group Bs each with 10 muons at the start.
According to your arithmetic both group Bs will have have lost 5 muons after 2.2 microseconds ie two groups of type B have lost 10 muons in total.
So two group Bs do exactly the same as one group A.