Question

1. 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

1. identify each as a permutation

Explanation:

Step1: Calculate the proportion of students who took Form B

Since half of the class took Form B, the proportion of students who took Form B is \(0.5\).

Step2: Calculate the proportion of students who did not pass Form B

If \(39\% = 0.39\) of the students who took Form B passed, then the proportion of students who did not pass Form B is \(1 - 0.39=0.61\).

Step3: Calculate the probability of a student taking Form B and not passing

Using the multiplication rule for independent - like events (since the event of taking Form B and the event of not passing (given Form B) are related in a multiplicative way here), the probability \(P\) that a student took Form B and did not pass is \(P = 0.5\times0.61\).

$$P=0.5\times(1 - 0.39)=0.5\times0.61 = 0.305$$

Answer:

C. \(0.305\)