Conditional Probability

Front

What is Conditional Probability?

Back

Conditional Probability is the probability of an event occurring given that another event has already occurred. It is denoted as P(A|B), which means the probability of event A occurring given that event B has occurred.

Tags

CCSS.HSS.CP.A.5

CCSS.HSS.CP.A.3

Front

How do you calculate Conditional Probability?

Back

Conditional Probability can be calculated using the formula: P(A|B) = P(A and B) / P(B), where P(A and B) is the probability of both events occurring.

Tags

CCSS.HSS.CP.A.3

CCSS.HSS.CP.A.5

Front

What is the formula for the probability of two independent events A and B occurring?

Back

For independent events, the probability of both A and B occurring is P(A and B) = P(A) * P(B).

Tags

CCSS.HSS.CP.A.2

CCSS.HSS.CP.A.4

Front

What is the difference between independent and dependent events?

Back

Independent events are those whose outcomes do not affect each other, while dependent events are those where the outcome of one event affects the outcome of another.

Tags

CCSS.HSS.CP.A.5

Front

If P(A) = 0.5 and P(B) = 0.4, what is P(A and B) if A and B are independent?

Back

P(A and B) = P(A) * P(B) = 0.5 * 0.4 = 0.2.

Tags

CCSS.HSS.CP.A.2

CCSS.HSS.CP.A.4

Front

What does it mean if P(B|A) = 0.3?

Back

It means that the probability of event B occurring given that event A has occurred is 30%.

Tags

CCSS.HSS.CP.A.3

CCSS.HSS.CP.A.5

Front

In a jar with 5 black and 3 white marbles, what is the probability of drawing a black marble first?

Back

P(Black) = Number of Black Marbles / Total Marbles = 5 / (5 + 3) = 5/8.

Tags

CCSS.7.SP.C.7B

Front

What is the probability of drawing a white marble second, given that the first marble drawn was black?

Back

If the first marble drawn was black, there are now 4 black and 3 white marbles left. P(White|Black first) = 3 / (4 + 3) = 3/7.

Tags

CCSS.HSS.CP.B.6

Front

What is the Law of Total Probability?

Back

The Law of Total Probability states that if B1, B2, ..., Bn are mutually exclusive events that cover the entire sample space, then P(A) = P(A|B1)P(B1) + P(A|B2)P(B2) + ... + P(A|Bn)P(Bn).

Front

If the probability of passing a test is 0.8 and the probability of studying is 0.9, what is the probability of passing given that the student studies?

Back

Using the formula P(Pass|Study) = P(Pass and Study) / P(Study). If P(Pass and Study) = 0.8, then P(Pass|Study) = 0.8 / 0.9 = 0.888.

Tags

CCSS.HSS.CP.A.3

CCSS.HSS.CP.A.5