at least 3 days in a row during a 5 day
3days -5* (1/2)^5=5/32

4 days-2*(1/2)^5=2/32

5days -1*(1/2)^5=1/32

5/32+2/32+1/32=8/32=1/4

my take follows >>
P(Rain at least 3 days)=P(Rain 3 days)+P(Rain 4 days)+P(Rain 5 days)
Since we have restriction "in a row" clue the rows such as 3,4,5 and find number of ways accordingly, 3C1, 2C1 and 1C1 or 3,2,and 1 ways with the same probability value counted as (1/2)^5

the tricky part is that when it says 3 days "in a row" it implies there can be "R-NR-R-R-R" and "R-R-R-NR-R" too, which makes additional 2 ways

Thus, P(Rain at least 3 days)= (1/2)^5 *(3+2+1+2)= 8/32 = 1/4