\[C(10, 2) = rac{10!}{2!(10-2)!} = rac{10 imes 9}{2 imes 1} = 45\] Next, we need to calculate the number of combinations where at least one item is defective. It’s easier to calculate the opposite (i.e., no defective items) and subtract it from the total.
The number of combinations with no defective items (i.e., both items are non-defective) is:
The number of non-defective items is \(10 - 4 = 6\) .
\[C(10, 2) = rac{10!}{2!(10-2)!} = rac{10 imes 9}{2 imes 1} = 45\] Next, we need to calculate the number of combinations where at least one item is defective. It’s easier to calculate the opposite (i.e., no defective items) and subtract it from the total.
The number of combinations with no defective items (i.e., both items are non-defective) is: probability and statistics 6 hackerrank solution
The number of non-defective items is \(10 - 4 = 6\) . \[C(10, 2) = rac{10