Question

half of a class took form a of a test, and half took form b. of the students who took form b, 39% passed. what is the probability that a randomly chosen student took form b and did not pass?
a 0.055
b 0.195
c 0.305
d 0.390

Explanation:

Step1: Calculate the probability of taking Form B

Since half of the class took Form B, the probability of taking Form B is \(P(\text{Form B})=\frac{1}{2} = 0.5\).

Step2: Calculate the probability of not passing given Form B

If 39% passed Form B, then the probability of not passing given Form B is \(P(\text{not pass}|\text{Form B})=1 - 0.39=0.61\).

Step3: Use the multiplication rule for conditional probability

The multiplication rule for conditional probability is \(P(A\cap B)=P(A)\times P(B|A)\). Here \(A\) is the event of taking Form B and \(B\) is the event of not passing. So \(P(\text{Form B}\cap\text{not pass})=P(\text{Form B})\times P(\text{not pass}|\text{Form B})\).
Substitute the values: \(P(\text{Form B}\cap\text{not pass})=0.5\times0.61 = 0.305\).

Answer:

C. \(0.305\)