Dependent and Independent Events
Independent Events
Given multiple events, independent events are when one event does not affect the outcome of another. Given independent events \(A\) and \(B\), we use \(A \cup B\) to denote both events A and B occuring.
To calculate the probability of \(A\) and \(B\) occurring, we use the following formula:
\(P(A \cap B) = P(A) \times P(B)\)
ExampleEmma has a regular deck of 52 cards. She randomly picks one, observes it, returns it to the deck, then does it again. What is the probability that she picks a jack card, and picks a spade card in these separate events?
First, we can let \(J\) be the event a jack card is picked and \(S\) be the event a spade is picked.
Next, we can calculate the probability of both a jack card and a spade card being picked as such:
\(P(J \cap S) = P(J) \times P(S)\)
\(P(J \cap S) = \cfrac{4}{52} \times \cfrac{13}{52}\)
\(P(J \cap S) = 0.01923\)
Therefore, we can determine the probability of Emma picking a jack card, return it, then pick a spade card is \(\boldsymbol{0.01923}\) (or \(\boldsymbol{1.923 \%}\)).
First, we can let \(T\) denote the event that the coin lands on tails, and \(5\) denotes the event that the die rolls a \(5\).
Next, we can calculate the probability of both the coin landing on tails and the die rolling a \(5\) as such:
\(P(T \cap 5) = P(T) \times P(5)\)
\(P(T \cap 5) = \cfrac{1}{2} \times \cfrac{1}{6}\)
\(P(T \cap 5) = \cfrac{1}{12} = 0.0833\)
Therefore, we can determine the probability that the coin lands on tails and the die rolls a \(5\) for Felix is \(\boldsymbol{0.08333}\) (or \(\boldsymbol{8.333\%}\)).
Dependent Events
A dependent even is an event which is affected by another.
Given two events, \(A\) and \(B\), the event that \(B\) occurs given that \(A\) has occurred is denoted by \(B|A\).
To calculate the probability of independent event A occurring and dependent event B occurring is as follows:
\(P(A \cap B)=P(A) \times P(B|A)\)
ExampleGertrude has a bag of marbles. \(3\) of which are blue and \(5\) of which are red. She picks one out of the bag and then another without returning the other. What is the probability of her picking a blue marble and then a red?
First, we can let \(B\) denote the event that a blue marble is removed and \(R\) denote the event that a red marble is removed.
Next, we can represent the probability that a blue marble is picked first as such:
\(P(B) = \cfrac{3}{8}\)
Then, we can represent the probability that a red marble is picked given a blue marble was picked first as such:
\(P(R|B) = \cfrac{5}{7}\)
Note how the denominator of P(R|B) is smaller because the first red marble hasn’t been returned to the bag.
Now we can calculate the probability of Gertrude picking a blue marble then a red marble by finding the product of the probabilities we calculated above:
\(P(B \cap R) = P(B) \times P(R|B)\)
\(P(B \cap R) = \cfrac{3}{8} \times \cfrac{5}{7}\)
\(P(B \cap R) = \cfrac{15}{56} = 0.2679\)
Therefore, we can determine the probability of Gertrude picking a blue marble then a red is \(\boldsymbol{0.2679}\) (or \(\boldsymbol{26.79\%}\)).
First, we can let \(D\) denote the event that a diamond is picked.
Next, we can represent the probability that a diamond is removed from a full deck as such:
\(P(D) = \cfrac{13}{52}\)
Then, we can represent the probability that a second diamond is picked without returning the first as such:
\(P(D|D) = \cfrac{12}{51}\)
Now, we can determine the probability of picking two diamonds as such:
\(P(D \cap D) = P(D) \times P(R|B)\)
\(P(D \cap D) = \cfrac{13}{52} \times \cfrac{12}{51}\)
\(P(D \cap D) = \cfrac{156}{2652} = 0.5882\)
Therefore, we can determine the chances of Harrison picking two diamonds from a deck of cards is \(\boldsymbol{0.05882}\) (or \(\boldsymbol{5.882\%}\)).