In a school, 14% of students take drama and computer classes, and 67% take drama Class. What is the probabilty that a students takes computer class given that the student takes drama class? i have no idea how to work this so please show me your work. Thanks!!
What is the probability that a student neither takes computer class nor takes drama class
To find the probability that a student takes a computer class given that the student takes a drama class, we can use conditional probability.
Let's define the following events:
A: Student takes drama class.
B: Student takes computer class.
We need to find P(B|A), which represents the probability of event B occurring given that event A has occurred.
We know that P(A) represents the probability of event A occurring, and P(B) represents the probability of event B occurring.
From the given information:
P(A) = 67% (or 0.67)
P(A and B) = 14% (or 0.14)
Now, we can use the formula for conditional probability:
P(B|A) = P(A and B)/P(A)
Substituting the given values:
P(B|A) = P(A and B)/P(A) = 0.14/0.67 ≈ 0.208 (rounded to three decimal places)
Therefore, the probability that a student takes a computer class given that the student takes a drama class is approximately 0.208.
Sure! To find the probability that a student takes computer class given that the student takes drama class, we can use conditional probability.
Let's denote:
A = the event that a student takes computer class
B = the event that a student takes drama class
We are given that 14% of students take drama and computer classes, so P(A ∩ B) = 0.14.
We are also given that 67% of students take drama class, so P(B) = 0.67.
The probability of A given B, denoted P(A|B), can be calculated using the formula:
P(A|B) = P(A ∩ B) / P(B)
Substituting the given values, we have:
P(A|B) = 0.14 / 0.67
Calculating this, we get:
P(A|B) ≈ 0.208 (rounded to three decimal places)
Therefore, the probability that a student takes computer class given that the student takes drama class is approximately 0.208, or 20.8%.
I hope this explanation helps! Let me know if you have any more questions.
This is a problem in "conditional probability".
The probability of E, given F is:
P(E/F) = P (E and F)/ P(F)
Call P(F) = drama = .67 given
Call P(E) = Computer = unknown
the P(EandF) =.14 given
Now the probability that a student takes computer given drama
= .14/.67 = .21