I write a two-digit number by randomly picking each digit from the set two, four, six, nine, where digits can be repeated. What is the probability that the sum of the two digits is greater than five?
If we’re trying to create a two-digit number and there are four options for the first digit and the same four options for the second digit by the basic counting principle. Which tells us if there are 𝑚 ways to do one thing and 𝑛 ways to do another, then there will be 𝑚 times 𝑛 ways to do both. We know there are four ways to choose the first digit and four ways to choose the second digit.
We have then 16 total possible outcomes. We know that probability equals the successful outcomes over total possible outcomes. For us, the outcome we’re considering is when the sum of the two digits is greater than five. We could use a grid like this to consider the sum of the two digits. In the first column and the first row, it would be the sum of two and two, which is four. Then we would consider if the first digit was two and the second digit was four, there would be a sum of six between the two digits.
We fill in the whole table in the same way. And if we look carefully, we see that only one of these values is less than five, which means the other 15 are greater than five. So we can say that the probability that two digits sum together are greater than five can happen 15 ways out of the 16 total options. 15 over 16 cannot be reduced. So the probability here is 15 16ths.