Getting a defective item when two items are randomly selected from a box of two defective and three non-defective items
Let A - ''getting a defective item'', then "P(A)=1-P(\\neg A)"
"P(\\neg A)={\\frac m n}" , where n - number of options of taking two items out of five, m - number of options of taking none defective item
"P(A)=1-{\\frac {{ 3\\choose 2}} {{5 \\choose 2}}}=1-{\\frac 3 {10}}={\\frac 7 {10}}"
Comments
Leave a comment