Answer:
P(Topic A) = 0.91
P(Topic B) = 0.9163
The student should choose Topic B to maximize the probability of writing a good paper because P(Topic B)>P(Topic A).
Step-by-step explanation:
If Topic A is chosen, 2 books will be ordered (n=2)
If topic B is chosen, 4 books will be ordered (n=4)
Probability that a book arrives in time (p) = 0.7
Probability that a book does not arrive in time (q) = 1 - p = 1 - 0.7 = 0.3
We will <u>use the binomial distribution</u> to find out which topic should the student choose to maximize the probability of writing a good paper. The binomial distribution formula is:
P(X=x) = ⁿCˣ pˣ qⁿ⁻ˣ
where p = probability of success
q = probability of failure
n = total no. of trials
x = no. of successful trials
For Topic A, n = 2, p=0.7 and q=0.3. If topic A is chosen, the student will use at least half the books i.e. he will use either 1 or 2 books. So,
P(Topic A) = P(X=1) + P(X=2)
=²C₁ (0.7)¹(0.3)²⁻¹ + ²C₂ (0.7)²(0.3)²⁻²
= 0.42 + 0.49
P(Topic A) = 0.91
For topic B, n=4, p=0.7 and q=0.3. If topic B is chosen, the student will choose 2 or more books i.e. 2, 3 or 4 books.
P(Topic B) = P(X=2) + P(X=3) + P(X=4)
= ⁴C₂ (0.7)²(0.3)⁴⁻² + ⁴C₃ (0.7)³(0.3)⁴⁻³ + ⁴C₄ (0.7)⁴(0.3)⁴⁻⁴
= 0.2646 + 0.4116 + 0.2401
P(Topic B) = 0.9163
The student should choose Topic B to maximize the probability of writing a good paper because P(Topic B)>P(Topic A) as calculated above.
