How much 5% salt solution must be added to 500 mL of a 20% solution to get a 7% solution? (edited)
Solution to yesterday’s problem:
When dealing with a probability problem like this, the result will be fraction where the numerator is the number of occurrences which satisfy the condition, and the denominator is the number of possible occurrences. It is often helpful to consider these two values separately until the end of the problem.
First let’s look at the total number of possible occurrences. There are five books possible, and we are choosing three. This is a combinations problem because the order of the books does not matter. Drawing our blanks,
____ ____ ____
Any of the five books could go in the first blank, four books in the second blank, and three books in the third blank.
__5__ __4__ __3__
Multiplying these, there are 60 permutations of possible books we could pick. But the order doesn’t matter, so we need to divide by the number of permutations that represent one combination (eg. ABC and ACB are the same). This number is always the factorial of the number of blanks. Why? Because that’s how many ways the sequence can be reordered. For instance,
ABC = ACB = BAC = BCA = CAB = CBA
3! is 6; dividing 60 by 6 gives 10 combinations. So 10 is our denominator. Now we must find in how many of those 10 combinations are all the books fiction. Let’s call the fiction books A, B, and C; the nonfiction books will be D and E. Then the only combination where A, B, and C are present is ABC (we eliminated the ones with different orders). Finally, our probability is 1/10.