The AND rule
Sometimes in life we want to know the probability of many things happening, not just one.
Example
Sarah has a bag containing 4 red balls and 5 green balls (9 balls in total). If she were to select a ball at random from the bag:
- What is the probability that she selects a red ball?
- What is the probability she selects a green ball?
Solution
- To solve this we realise that there are 4 red balls, out of 9 balls in total, leaving us with a probability of \(\frac{4}{9}\).
- There are 5 green balls out of a total of 9 balls so the probability is \(\frac{5}{9}\).
Question
Tim has a bag containing 3 white, 2 yellow and 4 red balls, what is the probability of him selecting:
- A white ball?
- A yellow ball?
- There are 3 white balls out of a total of 9 balls so the probability is \(\frac{3}{9}\).
- There are 2 yellow balls out of 9 balls in total so we have a probability of \(\frac{2}{9}\).
What if Tim selects 2 balls from the bag and we want to know the probability of him getting, for example, a white ball and then a red ball? First we need to know whether Tim is going to replace the ball after he selects it - we need to know this because if he doesn’t, then his selection for the first ball will affect the probabilities of the other balls.
For the two selections to be independent events, the first ball chosen must be replaced. The probability of him selecting a white ball the first time is \(\frac{3}{9}\). This ball is then replaced in the bag. The probability of him then selecting a red ball would be \(\frac{4}{9}\). In order to answer the question we need to find the probability he selects a white ball first and a red ball second. To do this we multiply the two probabilities.
P(white then red) = \(\frac{3}{9} \times \frac{4}{9} = \frac{12}{81} = \frac{4}{27}\)
Question
James is going to roll a dice and toss a coin. What is the probability that he gets:
- A 5 on the dice
- A 5 on the dice and a tail on the coin
- An even number and a head
- As there are 6 numbers on a dice but only one of them is the number 5 P(5) = \(\frac{1}{6}\)
- From 1) we know the probability of getting a 5 is \(\frac{1}{6}\). The probability of getting tails is \(\frac{1}{2}\). Using the AND rule, the probability of getting a 5 and a tail is \(\frac{1}{6} \times \frac{1}{2} = \frac{1}{12}\)
- There are three even numbers on a dice: 2, 4 and 6. As there are 6 numbers in total the probability of getting an even number is P(even) = \(\frac{3}{6}\). Using the AND rule, the probability of getting an even number and a head is \(\frac{3}{6} \times \frac{1}{2} = \frac{3}{12}\) which can be simplified to \(\frac{1}{4}\)