Question

A student can enter a course either as a beginner (73%) or as a transferring student (27%). It is found that 62% of beginners eventually graduate, and that 78% of transfers eventually graduate. Find:

1. the probability that a randomly chosen student is a beginner who will eventually graduate,

2. the probability that a randomly chosen student will eventually graduate,

3. the probability that a randomly chosen student is either a beginner or will eventually graduate, or both.

4. Are the events ‘eventually graduate’ and ‘enters as a transferring student’ statistically independent?

5. If a student eventually graduates, what is the probability that the student entered as a transferring student?

6. If two entering students are chosen as random, what is the probability that not only do they enter in the same way but that they also both graduate or fail?

Answer 1

Solution

Back-up Theory

For 2 events, A and B,

P(A ∪ B) = P(A) + P(B) - P(A ∩ B), in general and ………………………………..………(1)

A and B are independent if P(A ∩ B) = P(A) x P(B) ……………………………………… (1a)

If A and B are two events such that probability of B is influenced by occurrence or otherwise of A, then

Conditional Probability of B given A, denoted by P(B/A) = P(B ∩ A)/P(A)……....….(2)

(1) => P(B ∩ A) = P(B/A) x P(A) or P(A ∩ B) = P(A/B) x P(B) …………………………..(2a)

P(B) = {P(B/A) x P(A)} + {P(B/A^C) x P(A^C)}……………………………………………….(3)

P(A/B) = P(B/A) x {P(A)/P(B)}……………………………..…………………..…….(4)

Now to work out the solution,

Let A represent the event that a student enters a course as a beginner and B represent the event that the student eventually graduates

Then, trivially, A^C represents the event that the student does not enter a course as a beginner, i.e., he enters a course as a transferring student and B^C represents the event that the student eventually does not graduate. ………………………………………………………………………………….. (5)

Given data translates in probability language as follows:

P(A) = 0.73 (73%) and hence P(A^C) = 0.27 (27%) ………………………………………….. (6)

62% of beginners eventually graduate =>

P(B/A) = 0.62 and hence P(B^C/A) = 0.38 ……………………………………………………… (7)

78% of transfers eventually graduate =>

P(B/A^C) = 0.78 and hence P(B^C/A^C) = 0.22 …………………………………………………….. (8)

Part (a)

1. Probability that a randomly chosen student is a beginner who will eventually graduate
2. = P(A ∩ B)
3. = P(B ∩ A)
4. = 0.62 x 0.73 [vide (2a), (7) and (6)]
5. = 0.4526   Answer 1

Part (b)

6. Probability that a randomly chosen student will eventually graduate,

= P(B)

= {P(B/A) x P(A)} + {P(B/A^C) x P(A^C)} [vide (3)]

= (0.62 x 0.73) + (0.78 x 0.27) [(7), (6), and (8)]

= 0.4526 + 0.2106

= 0.6632 Answer 2

Part (c)

7. Probability that a randomly chosen student is either a beginner or will eventually graduate, or both.

= P(A ∪ B)

= P(A) + P(B) - P(A ∩ B) [vide (1)]

= 0.73 + 0.6632 – 0.4526 [vide (6), Answers 2 and 1]

= 0.9406   Answer 3

Part (d)

P(‘eventually graduate’)

= P(B)

= 0.6632 [vide Answer 2]

P(‘enters as a transferring student’)

= P(A^C)

= 0.27 [vide (6)]

Now, P(B ∩ A^C)

= P(B/A^C) x P(A^C) [vide (2a)]

= 0.78 x 0.27 [vide (8) and (6)]

= 0.2106

P(B) x P(A^C)

= 0.6632 x 0.27 [as above]

= 0.1791

Clearly, P(B ∩ A^C) ≠ P(B) x P(A^C)

So, vide (1a), B and A^C are not independent.

8. i.e., Events ‘eventually graduate’ and ‘enters as a transferring student’ are
9. NOT statistically independent Answer 4

Part (e)

10. If a student eventually graduates, probability that the student entered as a transferring student
11. = P(A^C/B)
12. = P(B/A^C) x P(A^C)/P(B) [vide (4)]
13. = (0.78 x 0.27)/0.6632 [vide (8), (6) and Answer 2]
14. = 0.3176 Answer 5

Part (f)

If two entering students are chosen at random, probability that they enter in the same way

P(Both enter as beginner or both enter as transferring students)

= P(Both enter as beginner) + P(Both enter as transferring students) [note that these two events are mutually exclusive]

= (0.73²) + (0.27²) [vide (6)]

= 0.6058

P(they both graduate or fail)

= P(they both graduate) + P(they both fail) [note that these two events are mutually exclusive]

= (0.62 x 0.78) + (0.28 x 0.22) [(7) and (8)]

= 0.5452

Thus, probability that not only do they enter in the same way but that they also both graduate or fail

= 0.6058 x 0.5452

= 0.3303 Answer

DONE