There are
ways of selecting two of the six blocks at random. The probability that one of them contains an error is
So
has probability mass function
These are the only two cases since there is only one error known to exist in the code; any two blocks of code chosen at random must either contain the error or not.
The expected value of finding an error is then
It has to be the third option 8(1+6) because 8+48= 56 and 8(1+6)= 56
A=12
120% of a =80% of b
120%a = 80%b
120a/100 = 80b/100
Divide through by 100
120a = 80b
Divide through by 80
120a/80 = b
But a=12
120x 12/80 = b
18 = b
b = 18
a+b = 12 + 18 = 30
Total fruits→120=4u
apples=30=1u
3u=90→banana and oranges
banana→1u
oranges→2u
oranges=60