Probability And Statistics 6 Hackerrank Solution Today

The number of non-defective items is \(10 - 4 = 6\) .

For our problem:

\[P( ext{no defective}) = rac{C(6, 2)}{C(10, 2)} = rac{15}{45} = rac{1}{3}\] probability and statistics 6 hackerrank solution

The final answer is:

\[P( ext{at least one defective}) = 1 - rac{1}{3} = rac{2}{3}\] Here’s a Python code snippet that calculates the probability: 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: The number of combinations with no defective items (i

\[C(6, 2) = rac{6!}{2!(6-2)!} = rac{6 imes 5}{2 imes 1} = 15\] Now, we can calculate the probability that at least one item is defective: