b If two events A and B are dependent, then the probability that A and B will occur is P(AandB)=P(A)* P(B|A).
A
a 17/32 or ≈ 0.531
B
b 1/31 or ≈ 0.032
Practice makes perfect
a We have a bag that contains all of the possible chess pieces in a set. The table shows how many of each type of chess piece in each color there is.
King
Queen
Bishop
Rook
Knight
Pawn
Black
1
1
2
2
2
8
White
1
1
2
2
2
8
We choose one piece at random and want to find the probability that we draw a black piece or a queen. Let event A be choosing a black piece and event B be selecting a queen. We are interested the following probability.
P( A or B)
Notice that a queen can be black. Therefore, events A and B are overlapping as they have common outcomes. In order to obtain P(A or B), we can use the following formula.
P(AorB)
=
P(A)+P(B)-P(AandB)
We need to find the probabilities on the right-hand side of this equation. We can calculate P( A) using the theoretical probability. We need to compare the number of favorable outcomes to the number of possible outcomes.
P( A) = favorable outcomes/possible outcomes
To find the number of favorable outcomes, we need to find the total number of black pieces. Let's add the numbers from the first row of the table to find this total.
King
Queen
Bishop
Rook
Knight
Pawn
Sum
Black
1
1
2
2
2
8
16
White
1
1
2
2
2
8
The number of favorable outcomes is 16. Since there are exactly as many white pieces as black pieces, we can find the number of possible outcomes by doubling the number of black pieces.
2 * 16 = 32
We are ready to calculate P(A).
From the table, we also know that there is 1 black queen and 1 white queen. With this information, we can conclude that the number of favorable outcomes for event B is 2. As we found P( A), we can find P( B) in the same way.
The last probability we need to find is P(AandB). Since there is only one queen that is black there is only 1 favorable outcome.
P(A and B) = 1/32
We can substitute the obtained probabilities into the formula for P(AorB) and simplify.
Finally, we found the probability that a randomly selected piece is black or a queen is 1732 or about 0.531.
b This time, when we randomly select pieces from the bag, we do not replace the first piece before drawing the second. Let the event of choosing a king be A and let choosing a pawn be B. We want to find the following probability.
P( A and B)
Since we do not replace the first piece before we select the second one, the occurrence of the first event affects the occurrence of the second. Therefore, these are dependent events.
Probability of Dependent Events
If two events A and B are dependent, then the probability that A and B will occur is
P(AandB)=P(A)* P(B|A).
We need to find the probabilities on the right-hand side of this equation. We can calculate P( A) using the theoretical probability just as we did in Part A.
P( A) = favorable outcomes/possible outcomes
The number of possible outcomes is still 32. To find the number of favorable outcomes, we need to find the number of kings in the bag. Let's analyze the table once again.
King
Queen
Bishop
Rook
Knight
Pawn
Black
1
1
2
2
2
8
White
1
1
2
2
2
8
We can see there are 2 kings in the bag. We are ready to calculate P(A).
Now, we can calculate P(B|A). The number of possible outcomes in the second drawing is different than in the first one, because we do not replace the first piece.
32-1= 31
Since in the first drawing we selected the king, there are still 8+8= 16 pawns that could be chosen. We have enough information to calculate P(B|A).
