Difference between revisions of "Conditional probability"

(Different Problems)
(Different Problem)
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Solution: The probability that the sum of the first two die is equal to the 3rd die is just another way to also equivalent to the probability that the sum of the numbers on the first two die is less than 7, which is <math>\frac {5}{12}</math>. The probability that a two is rolled and that it meets the first condition is <math>\frac {2}{9}</math>. dividing gets us <math>\frac {8}{15}</math>.
 
Solution: The probability that the sum of the first two die is equal to the 3rd die is just another way to also equivalent to the probability that the sum of the numbers on the first two die is less than 7, which is <math>\frac {5}{12}</math>. The probability that a two is rolled and that it meets the first condition is <math>\frac {2}{9}</math>. dividing gets us <math>\frac {8}{15}</math>.
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==Review Problems==
  
 
== See also ==
 
== See also ==

Revision as of 17:58, 30 June 2024

Conditional probability is the probability of an event occurring, assuming that another event has already occurred. $P(B|A)$ is said as the probability of event B given A


Example

Let us say that 2 fair 6 sided dice are rolled and their face up values sum is 6. What is the probability that the face up value of the one dice is 2?

Solution

Let call the first dice $D_1$ and the second one $D_2$. There are 5 ways for $D_1 + D_2 = 6$ and 2 of those ways (distinct) includes a 2. Therefore, our answer is $\frac {2} {5}$.

Formula

The formula for conditional probability is \[P(A \cap B) = P(A) \cdot P(B|A)\] where $P(B|A)$ represents the conditional probability. $P(B|A)$ is also said as the probability of event B occurring given event A occurs. $P(A \cap B)$ is the probability $P(A) \cdot P(B)$. We can also represent $P(B|A)$ as \[P(B|A) = \frac {P(A \cap B)} {P(A)}\]

Different Problem

A fair standard die is tossed 3 times. Given that the sum of the first two tosses equals the third, what is the probability that at least one 2 is tossed? (Source AMC)


Solution: The probability that the sum of the first two die is equal to the 3rd die is just another way to also equivalent to the probability that the sum of the numbers on the first two die is less than 7, which is $\frac {5}{12}$. The probability that a two is rolled and that it meets the first condition is $\frac {2}{9}$. dividing gets us $\frac {8}{15}$.

Review Problems

See also