At the end of a training programme, students have to pass an exam to gain a cirtificate.
The probability of passing the exam at the first attempt is 0.75
Those who fail are allowed to re-sit.
The probability of passing the re-sit is 0.6
No further attempts are allowed.
(a)(i) Create a tree diagram.
(ii)What is the probability that a student fails to gain a certificate?
(b) Three students take the exam.
What is the probability that all of them gain a certificate?
+23246 a) A tree diagram is hard to explain in words, but I'm not sufficiently computer literate to provide a drawing.
So: let the diagram start at point A.
From A, there are two paths, one is a successful path to B (with value 0.75) and the other is an unsuccessful path to C (with value 0.25).
There is no path from B because there is no need to take another test.
From point C, there are two paths, one is a successful path to D (with value 0.60) and the other is an unsuccessful path to E (with value 0.40)
There are no paths from D or E.
b) The only way for a student to fail is to go from A to C (0.25) and from C to E (0.40). To find the probability for going from A to E is 0.25 x 0.40 = 0.10.
c) Since the protability of failing is 0.10, the probability of succeeding is 0.90.
The probability of all succeeding is 0.90 x 0.90 x 0.90 = 0.729.
