Probability (Multiple independent events: And)
When you want to find the probability of multiple events happening, your probability that ALL events will happen is always going to be smaller than the probability that only one of the events will happen. So, whenever you have an AND probability, you want to combine the probability of each individual event into one total probability (which will always be smaller than the smallest individual probability).
When you want to combine probabilities to account for the probability of ALL things happening (an event AND another event), you multiply probabilities. (This may seem strange at first, but remember: probabilities are fractions and when you multiply a number by a fraction, it gets smaller!)
Let's stick with our example: the probability of getting any particular number when you roll a die is $\dfrac{1}{6}.
Example:
Find the probability of getting a 3 on one die AND a 3 on a second die.
To find the probability of BOTH events, you multiply the probabilities of each event together.
The probability of getting a 3 on one die is $\dfrac{1}{6}$.
The probability of getting a 3 on the other die is $\dfrac{1}{6}$.
Multiply the probability of each together to find the probability of BOTH events happening:
$\dfrac{1}{6} \times \dfrac {1}{6} = \dfrac{1}{36}$
Your probability of getting a 3 AND a 3 is $\dfrac{1}{36}$.
If you wanted to add in the probability of another event happening, you'd just multiply in that probability as well.
Example:
Find the probability of getting a 3 on one die AND a 3 on a second die AND "heads" when you flip a coin.
To find the probability of ALL events happening, you multiply the probabilities of each event together.
The probability of getting a 3 on one die is $\dfrac{1}{6}$.
The probability of getting a 3 on the other die is $\dfrac{1}{6}$.
The probability of getting "heads" on a coin flip is $\dfrac{1}{2}$.
Multiply the probability of each together to find the probability of ALL events happening:
$\dfrac{1}{6} \times \dfrac {1}{6} \times \dfrac{1}{2}= \dfrac{1}{72}$
Your probability of getting a 3 AND a 3 and "heads" is $\dfrac{1}{72}$.
Overall, anytime you want to find the probability of multiple events happening, find the probability of each event, and multiply the probabilities together. Remember, your final probability should be lower than any of the individual probabilities!
Practice Problems:
Probability (Multiple independent events: AND)
Calculate the probability of the following sets of events:
You have cupboard full of plates, but you do not have a full set of any dishes. You have 3 blue plates, 6 flowered plates, 2 white plates and 1 green plate.
- If you are eating alone, and you choose randomly, what is the probability that you get the green plate?
- If you are eating with one other person, what's the probability that you get the green plate and a blue plate?
- Or, if you prefer matching plates, what's the probability that you pull out two blue plates?
- If you choose randomly, what is the probability that you will pull out 4 flowered plates in a row?
- Let's say you hate the flowered plates, what's the probability that you pull 2 blue plates and 2 white plates?
- What's the probability of getting 4 non-flowered plates?
You are playing cards with a friend. There are 52 cards in a standard deck, divided into 4 suits (13 hearts, 13 spades, 13 clubs, 13 diamonds). Each suit contains an ace, king, queen, jack, 10, 9, 8, 7, 6, 5, 4, 3, and 2 (there are 4 of each of these cards in each deck).
- What is the probability that you get a pair of aces in your first two cards?
- What is the probability that you get 3 kings in the first three cards?
- What is the probability that you get a flush, all 5 of the cards in your hand are the same suit?
- What is the probability that you get a royal straight (specfically: get the following cards in this order:10, Jack, Queen, King, Ace)?
- What is the probability that you get a full house (specifically, first three of a kind and then a pair)?
- What is the probability that you get four kings and an ace (in that order)?
Answer Key: