Independent and Dependent Events

We need to understand independent and dependent events to be able to do the next sections.

• Two or more events are independent if one event doesn’t effect the probability of the others happening.
• Two or more events are dependent if one event does effect the probability of the others happening.

Example:

• Getting a head both times on 2 coin flips are independent events.
• Picking a red marble at random from a bag, then picking a green marble without replacing the red marble are dependent events. 

The AND Rule

• The AND rule states that: If two events, A and B, are independent, then, 

\text{P}(A \text{ and } B) = \text{P}(A) \times \text{P}(B)

This means to find the probability of A and B occurring you must multiply the probability of A occurring by the probability of B occurring.

The OR Rule

• The OR rule states that: for two events, A and B, then,  

\text{P}(A \text{ or } B) = \text{P}(A) + \text{P}(B) - \text{P}(A \text{ and } B)

• If A and B cannot happen together, we say they are mutually exclusive, and then we have \text{P}(A \text{ and } B) = 0, so the OR rule becomes

\text{P}(A \text{ or } B) = \text{P}(A) + \text{P}(B)

Probability Trees

Probability trees are similar to frequency trees, but we instead put the probabilities on the branches and the events at the end of the branch.

Example: A bag contains 4 red balls and 5 blue balls.

Raheem picks 2 balls at random.

Calculate the probability that he selects the same coloured ball each time, given that after each time a ball is selected, it is replaced.

Step 1: Construct the probability tree showing two selections.

We know there are a total of 9 balls in the bag so there is a \dfrac{4}{9} chance of picking a red ball.

Then as the red ball is replaced, there are still 4 red balls left out of 9, so again there is a  \dfrac{4}{9} chance of picking a red ball on the second selection.

Continue and fill in the rest.

Step 2: Use the AND rule

From the tree diagram we can see that there are two ways of doing this, either

blue and blue, or red and red

We use the AND rule via the tree diagram, so

\text{P(blue and blue)}=\dfrac{5}{9}\times\dfrac{5}{9}= \textcolor{blue}{\dfrac{25}{81}} \,\, and \,\, \text{P(red and red)}=\dfrac{4}{9}\times\dfrac{4}{9}= \textcolor{red}{\dfrac{16}{81}}

Step 3: Use the OR rule

The final step then is to add the probabilities together, by the OR rule for mutually exclusive events, to get,

\text{P(same colour)}= \dfrac{25}{81} +\dfrac{16}{81}=\dfrac{41}{81}

Conditional Probability

The conditional probability of A given B, is the “probability that event A happens given that event B happens”. You will not be told that it is a conditional probability question, but seeing words like ‘without replacement’ or ‘given’ will mean that it is one, or you may have to use your own intuition.

If two events, A and B, are independent, then

\textcolor{black}{\text{P}(A \text{ given } B) = \text{P}(A)} \,\,     and    \,\, \textcolor{black}{\text{P}(B \text{ given } A) = \text{P}(B)}

If two events, A and B are dependent, then

\textcolor{black}{\text{P}(A \text{ and } B) = \text{P}(A) \times \text{P}(B \text{ given } A)}

Example:

Benjamin plays football for his local team. The probability that he is in the starting line up for his team this Sunday is 0.7. If he starts the game, the probability that he scores a goal is 0.4.

What is the probability that Benjamin starts the game but doesn’t score a goal?

Step 1: We want to find \text{P(starts and doesn't score)}

Let “starts” be event A and “doesn’t score” be event B

Step 2: \text{P}(A) = 0.7

\text{P}(B \text{ given } A) = \text{P(doesn't score given he starts)} = 1 - 0.4 = 0.6

Step 3: Then,

\text{P}(A \text{ and } B) = \text{P}(A) \times \text{P}(B \text{ given } A) = 0.7 \times 0.6 = 0.42

Conditional Probability Trees

Conditional probability trees are similar to probability trees, but the probabilities change depending on the previous events.

Example: A bag contains 4 red balls and 5 blue balls.

Raheem picks 2 balls at random.

Calculate the probability that he selects the same coloured ball each time, given that each time a ball is selected, it is not replaced.

Step 1: Construct the probability tree showing two selections,

There are 9 balls to begin with, reducing to 8 after the first selection, as shown below,

The chance of selecting a red ball for the first selection is \dfrac{4}{9}, then with one red ball removed,  the second selection is \dfrac{3}{8} and so on….

Step 2: Use the tree diagram to determine the probability of selecting the same colour twice. We can see that there are two ways of doing this, either blue and blue, or red and red. We use the AND rule via the probability tree, so

\text{P(blue and blue)}=\dfrac{5}{9}\times\dfrac{4}{8}= \textcolor{blue}{\dfrac{20}{72}} \text{ and  } \text{P(red and red)}=\dfrac{4}{9}\times\dfrac{3}{8}= \textcolor{red}{\dfrac{12}{72}}

Step 3: Add the probabilities together, by the OR rule for mutually exclusive events, to get

\text{P(Same colour)}= \dfrac{20}{72} +\dfrac{12}{72}=\dfrac{32}{72}