Please see second method of arjun sir answer.
probability of accessing $M_1$ is $1$ since we always start searching data from memory $M_1$
$0.99$ gives probability that we will access $M_1$ and get the data i.e. hit in $M_1$
this means probability that we access $M_1$ but don't get data $=1 - M_1 = 1- 0.99$ i.e. we will now access $M_2$ and try to get data from it. This is miss in $M_1$
So probability that we access $M_2$ is $1-0.99$ i.e. when there is miss in $M_1$.
Probability that we access $M_2$ and get data is $0.00998$
this means probability that we access $M_2$ but don't get data $=1 - M_2 = 1- 0.00998$ i.e. we will now access $M_3$ and try to get data from it. This is miss in $M_2$ given that there is already a miss in $M_1$.
So now at last we access $M_3$ and we will definitely get the data so Hit in $M_3$ is 1 and miss in $M_3$ is $0$.
Also probability that we will access $M_3$ is $1-0.00998$ or $0.00002$ i.e. when there is miss in $M_2$
Now what to do when there is a miss is given in question.
Here they have given relative probability so thats why we don't need to write like $(1-0.99)(1-0.998).$