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Independent events

Suppose we need to calculate the probability of $2$ events happening. In the lesson of probability, we had examples of throwing two dice and calculating the probability that both are $5$. We know that the probability of getting $5$ on first die is equal to $\frac{1}{6}$. However, what happens with the second die? The dice aren’t connected and the probability of getting $5$ on second die is also $\frac{1}{6}$. That will not change no matter what number we get on the first die.

Or for example, we can flip a coin a few times. Let’s say we get heads $5$ times in a row. Some people would think the $6$th flip should give us tails, since we got heads so many times before. However, that’s not the case. The coin doesn’t know what were the previous outcomes and each flip is an isolated event. As a result, $6$th flip is just as likely to be heads or tails, no matter what the previous outcomes were.

Events like this, where the outcome of one of the events doesn’t affect the outcome of another are called independent events.

Multiplication rule 

Example 1
 When flipping a coin, what are the odds of getting two heads in a row?

Solution
All the possible outcomes when flipping a coin two times are: getting two heads in a row; getting heads and then tails; getting tails first and then heads and lastly, getting two tails in a row.  We could write it as a set $\{HH, HT, TH, TT\}$, where $H$ stands for heads and $T$ for tails. Total number of all possible outcomes is $4$ and all are equally possible.

Another way of interpreting is: for the first flip, we have $2$ possibilities, and for the second flip we also have $2$ possibilities. In total that is $2\cdot2 = 4 $ possibilities. That will be our denominator in formula for calculating probability.

Let $A$ be our wanted outcome, $A= \{HH\}$. That’s only $1$ outcome out of possible $4$ that meets our conditions. As a result, probability of $A$ is: $$\displaystyle{P(A)=\frac{1}{4}}.$$

However, since we know the events of flipping a coin $2$ times in a row are independent events we can think about this example a little differently. We can say that probability of getting two heads in a row equals to getting heads on the first coin flip times the probability of getting heads on the second coin flip.

$P(H$ and $\displaystyle{H)=P(H_{1})\cdot P(H_{2}) = \frac{1}{2} \cdot \frac{1}{2}= \frac{1}{4}}.$

 

In general, if we have independent events $A$ and $B$ and we want to find the probability of both occurring we use the multiplication rule for independent events

$P(A$ and $B)= P(A \cap B)= P(A) \cdot P(B).$

Examples

Example 2
Rob has $2$ different hats, $5$ different shirts, $4$ different pants and $6$ different socks in his closet. He has exactly one brown hat, exactly one yellow shirt, exactly one blue pants and exactly one white socks. If he randomly selects each clothing item, what is the probability that he will wear brown hat, yellow shirt, blue pants and white socks?

Solution
Since he has $2$ different hats, and only one is brown, the probability of selecting brown hat is $\frac{1}{2}$. A probability of him picking yellow shirt out of $5$ different ones is $\frac{1}{5}$. Following the same process, we get that probability of picking blue pants is equal to $\frac{1}{4}$ and probability of picking white socks is $\frac{1}{6}$.

These events are independent, the selection of a hat doesn’t affect the selection of a shirt. Additionally, the selection of hat and shirt doesn’t affect the selection of pants, and so on.  Therefore, we can apply multiplication rule to find out the probability of him picking the right items at the same time.

Wanted probability is equal to $$\displaystyle{\frac{1}{2} \cdot \frac{1}{5} \cdot \frac{1}{4} \cdot \frac{1}{6} = \frac{1}{240}}.$$

Example 3
A jar contains $5$ red, $2$ green, $4$ blue and $7$ yellow marbles. We pick one marble from a jar and then put it back in. Then we pick another marble. What is the probability of choosing a yellow and then a red marble?

Solution
Since we have $7$ yellow marbles in a jar of $18$ marbles, the probability of picking the yellow marble is $\frac{7}{18}$. After seeing the color of a marble, we put it back in the jar. Consequently, the jar still has $18$ marbles. The probability of picking a red marble second is $\frac{5}{18}$.

These two events are independent because we put back the first marble, which makes the conditions for the second pick the same as they were for a first pick. Using the multiplication rule, the probability of choosing a yellow and then a red marble is:

$P($yellow and red$)=P($yellow$) \cdot P($red$)$ $$=\displaystyle{\frac{7}{18} \cdot \frac{5}{18}}$$ $$=\displaystyle{\frac{35}{324}}$$

Important!
If we hadn’t put back the first marble, the events of picking yellow and then red marble wouldn’t be considered independent events. The second event would be affected by the first one since we would have one less marble in a jar. Events like that are considered dependent events.

Addition rule

As we’ve seen in previous lesson, addition formula is $$P(A \cup B)=P(A)+P(B)-P(A \cap B)$$ However, we can apply multiplication rule on the $P(A \cap B)$.
The addition formula for independent events is:$$P(A \cup B)=P(A)+P(B)-P(A)P(B)$$

Example 4
Lets draw a card from a well-shuffled standard deck of cards. Now we want to determine the probability of drawing an ace or a heart.

Solution
Lets put $A$ for “ace” and $H$ for “heart”. We want to determine $P(A \cup H)$. Firstly, lets find the probabilities of these two separate events.

Since there are $4$ aces in a deck of $52$ cards, the probability of picking one ace is $\frac{4}{52}$. There are $13$ hearts, so the probability of picking one is $\frac{13}{52}$. These events are not mutually exclusive, therefore their intersection isn’t empty. We can pick a card that is an ace and a heart, and there is only one card like that. The probability of picking it is $\frac{1}{52}$. Lets combine these probabilities using the generalized addition rule.

$P(A \cup H)=P(A)+P(H)-P(A \cap  H)$
$P(A \cup H)=\displaystyle{\frac{4}{52} +\frac{13}{52}-\frac{1}{52}=\frac{4}{13}}$

Remember:
$\bullet$ or  represents union
$\bullet$ and  represents intersection