Probability problems are prevalent and have varying degrees of difficulty. For example:
What is the probability that I roll an even number on a standard 6-sided die?
This problem is pretty simple. There are 3 even numbers on a 6-sided die: 2, 4, and 6. There are a total of 6 numbers on a die, so the probability is 3/6 or 1/2.
Let’s try one that is a little more complicated:
Ava flips a coin. She stops when she lands a tail. What is the probability that she lands a tail on one of her first two tries?
You might think that the answer is 1/2 + 1/2 = 1. However, this is not the case. If this were true, it would be guaranteed for her to roll a tail; however, there is a chance to roll two heads. So, how can we find the true solutions? There are two ways to solve this problem.
The first way to solve this is to do casework. The probability that she lands a tail on her first try is 1/2. The probability that she lands a tail on her second try is equal to the probability that she lands a head on the first try and a tail on the second. Thus, the probability is 1/2 x 1/2 = 1/4.
So, the total probability is 1/2 + 1/4 = 3/4.
You can also use complimentary counting to solve this. The probability that she doesn’t land a tail on one of her first two tries is equal to the probability that she lands a head on both, which is 1/2 x 1/2 = 1/4.
Thus, the total probability is 1 - 1/4, which is also 3/4.
Let’s do a harder problem:
Raj has a sack of marbles. In the sack, there are 3 red marbles and 4 blue marbles. After selecting a marble, he places it back in the sack. What is the probability that Raj selects two blue marbles before selecting a red marble?
This problem seems complicated, but it can be simplified using complementary counting. The probability of selecting a red marble first is 3/7. The probability of selecting a blue marble and then a red marble is 4/7 x 3/7 = 12/49. So, the total probability of selecting a red marble before two blue marbles is 3/7 + 12/49 = 21/49 + 12/49 = 33/49.
So, the probability of selecting two blue marbles before a red marble is 1 - 33/49, which equals 49/49 - 33/49, which is 16/49.