IB Mathematics SL 4.5 Concepts of trial, outcome, equally likely outcomes AI SL Paper 2- Exam Style Questions- New Syllabus

(a)
Complete the tree diagram with the given probabilities:
– Does not use banned substances: \( 0.94 \)
– Tests positive: \( 0.02 \)
– Tests negative: \( 0.98 \)
– Uses banned substances: \( 0.06 \)
– Tests positive: \( 0.71 \)
– Tests negative: \( 0.29 \)
(Note: \( 0.29 = 1 – 0.71 \), \( 0.02 = 1 – 0.98 \), \( 0.94 = 1 – 0.06 \))

(b)

1. (i)
   Multiply the probabilities along the “does not use and tests negative” branch:
   \( 0.94 \times 0.98 = 0.9212 \)
   The probability is \( 0.921 \) (or \( 92.1\% \)).
2. (ii)
   For two athletes, square the probability from (b)(i):
   \( (0.9212)^2 = 0.848609 \)
   The probability is \( 0.849 \) (or \( 84.9\% \)).

(c)

1. (i)
   Calculate the probability of an incorrect test result (false positive or false negative):
   \( 0.94 \times 0.02 + 0.06 \times 0.29 \)
   \( = 0.0188 + 0.0174 = 0.0362 \)
   The probability is \( 0.0362 \) (or \( 3.62\% \)).
2. (ii)
   Multiply the probability from (c)(i) by the sample size:
   \( 0.0362 \times 1300 = 47.06 \)
   The expected number is \( 47.1 \).

(d)
Since none use banned substances, the probability of testing positive is \( 0.02 \) for each athlete. Use the binomial probability for \( P(X = 0) \) with \( n = 20 \), \( p = 0.02 \):
\( P(X = 0) = (1 – 0.02)^{20} \)
\( = 0.98^{20} \approx 0.667607 \)
The probability that none test positive is \( 0.668 \) (or \( 0.667607 \)).

(e)
With \( p = 0.02 \), calculate \( P(X \geq 3) \) for \( n = 20 \):
\( P(X \geq 3) = 1 – [P(X = 0) + P(X = 1) + P(X = 2)] \)
\( P(X = 0) = 0.667607 \)
\( P(X = 1) = 20 \times 0.02 \times 0.98^{19} \approx 0.272478 \)
\( P(X = 2) = \frac{20 \times 19}{2} \times 0.02^2 \times 0.98^{18} \approx 0.052846 \)
\( P(X \geq 3) = 1 – (0.667607 + 0.272478 + 0.052846) \approx 0.007069 \)
The probability that more than 2 test positive is \( 0.00707 \) (or \( 0.00706869 \)).