a box contains 9 slips that are each labeled with one number : 1 , 2 , 3 , 5 , 8 , 13 , 21 , and 55 . two of the slips are drawn at random from the box without replacement . what is the probability that the sum of the numbers on the two slips is equal to one of the numbers left in the box ?
probability = no : of desired outcomes / total no : of outcomes . you are picking two slips out of 8 slips . so total no : of outcomes = 8 c 2 = 28 desired outcome : sum of the numbers on the two slips is equal to one of the numbers left in the box . how many such outcomes are there ? if you look at the numbers closely , you will see that the following pair of numbers will give you the desired outcome . ( 12 ) ( 23 ) ( 35 ) ( 58 ) ( 813 ) . there are 5 such pairs . if the two numbers which i pick is from any of these 5 pairs , then i get my desired outcome . so no : of desired outcomes = 5 probability = 5 / 28 answer : d
divide(5, add(add(multiply(3, 5), const_10), const_3))