IB Mathematics SL 4.8 Binomial distribution. Mean and variance-AI SL Paper 1- Exam Style Questions- New Syllabus
Question
Noah administers a polygraph test to determine whether people are telling the truth, but it is not completely accurate. When a person tells the truth, they have a 20% chance of failing the test. Each test outcome is independent of any previous test outcome.
10 people take a polygraph test, and all 10 tell the truth.
(a) Calculate the expected number of people who will pass this polygraph test. [2]
(b) Calculate the probability that exactly 4 people will fail this polygraph test. [2]
(c) Determine the probability that fewer than 7 people will pass this polygraph test. [3]
▶️ Answer/Explanation
Markscheme
(a)
Probability of failing: \( P(\text{fail}) = 0.2 \). Probability of passing: \( P(\text{pass}) = 1 – 0.2 = 0.8 \). Number of people: \( n = 10 \).
Expected number of passes:
\[ \begin{aligned} E[X] &= n \times p = 10 \times 0.8 = 8 \end{aligned} \]
Answer: \( 8 \, \text{people} \) M1 A1
[2 marks]
Probability of failing: \( P(\text{fail}) = 0.2 \). Probability of passing: \( P(\text{pass}) = 1 – 0.2 = 0.8 \). Number of people: \( n = 10 \).
Expected number of passes:
\[ \begin{aligned} E[X] &= n \times p = 10 \times 0.8 = 8 \end{aligned} \]
Answer: \( 8 \, \text{people} \) M1 A1
[2 marks]
(b)
Model failures as a binomial random variable: \( n = 10 \), \( p = 0.2 \), \( q = 0.8 \). Probability of exactly 4 failures:
\[ \begin{aligned} P(X = 4) &= C(10, 4) \times (0.2)^4 \times (0.8)^6 \\ C(10, 4) &= \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210 \\ (0.2)^4 &= 0.0016 \\ (0.8)^6 &= 0.262144 \\ P(X = 4) &= 210 \times 0.0016 \times 0.262144 \approx 210 \times 0.0004194304 \approx 0.088080384 \approx 0.0881 \end{aligned} \]
Answer: \( 0.0881 \) M1 A1
[2 marks]
Model failures as a binomial random variable: \( n = 10 \), \( p = 0.2 \), \( q = 0.8 \). Probability of exactly 4 failures:
\[ \begin{aligned} P(X = 4) &= C(10, 4) \times (0.2)^4 \times (0.8)^6 \\ C(10, 4) &= \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210 \\ (0.2)^4 &= 0.0016 \\ (0.8)^6 &= 0.262144 \\ P(X = 4) &= 210 \times 0.0016 \times 0.262144 \approx 210 \times 0.0004194304 \approx 0.088080384 \approx 0.0881 \end{aligned} \]
Answer: \( 0.0881 \) M1 A1
[2 marks]
(c)
Fewer than 7 passes means 4 or more failures (\( \text{passes} + \text{failures} = 10 \)). Let \( X \) be the number of failures: \( P(\text{passes} < 7) = P(X \geq 4) = 1 – P(X \leq 3) \).
Calculate \( P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) \):
\[ P(X = k) = C(10, k) \times (0.2)^k \times (0.8)^{10-k} \]
For \( k = 0 \): \( C(10, 0) = 1 \), \( (0.2)^0 = 1 \), \( (0.8)^{10} = 0.1073741824 \), \( P(X = 0) \approx 0.1073741824 \).
For \( k = 1 \): \( C(10, 1) = 10 \), \( (0.2)^1 = 0.2 \), \( (0.8)^9 = 0.134217728 \), \( P(X = 1) = 10 \times 0.2 \times 0.134217728 \approx 0.268435456 \).
For \( k = 2 \): \( C(10, 2) = 45 \), \( (0.2)^2 = 0.04 \), \( (0.8)^8 = 0.16777216 \), \( P(X = 2) = 45 \times 0.04 \times 0.16777216 \approx 0.301989888 \).
For \( k = 3 \): \( C(10, 3) = 120 \), \( (0.2)^3 = 0.008 \), \( (0.8)^7 = 0.2097152 \), \( P(X = 3) = 120 \times 0.008 \times 0.2097152 \approx 0.201326592 \).
Sum:
\[ P(X \leq 3) \approx 0.1073741824 + 0.268435456 + 0.301989888 + 0.201326592 \approx 0.8791261184 \]
\[ P(X \geq 4) = 1 – 0.8791261184 \approx 0.1208738816 \approx 0.121 \]
Answer: \( 0.121 \) A1 M1 A1
[3 marks]
Fewer than 7 passes means 4 or more failures (\( \text{passes} + \text{failures} = 10 \)). Let \( X \) be the number of failures: \( P(\text{passes} < 7) = P(X \geq 4) = 1 – P(X \leq 3) \).
Calculate \( P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) \):
\[ P(X = k) = C(10, k) \times (0.2)^k \times (0.8)^{10-k} \]
For \( k = 0 \): \( C(10, 0) = 1 \), \( (0.2)^0 = 1 \), \( (0.8)^{10} = 0.1073741824 \), \( P(X = 0) \approx 0.1073741824 \).
For \( k = 1 \): \( C(10, 1) = 10 \), \( (0.2)^1 = 0.2 \), \( (0.8)^9 = 0.134217728 \), \( P(X = 1) = 10 \times 0.2 \times 0.134217728 \approx 0.268435456 \).
For \( k = 2 \): \( C(10, 2) = 45 \), \( (0.2)^2 = 0.04 \), \( (0.8)^8 = 0.16777216 \), \( P(X = 2) = 45 \times 0.04 \times 0.16777216 \approx 0.301989888 \).
For \( k = 3 \): \( C(10, 3) = 120 \), \( (0.2)^3 = 0.008 \), \( (0.8)^7 = 0.2097152 \), \( P(X = 3) = 120 \times 0.008 \times 0.2097152 \approx 0.201326592 \).
Sum:
\[ P(X \leq 3) \approx 0.1073741824 + 0.268435456 + 0.301989888 + 0.201326592 \approx 0.8791261184 \]
\[ P(X \geq 4) = 1 – 0.8791261184 \approx 0.1208738816 \approx 0.121 \]
Answer: \( 0.121 \) A1 M1 A1
[3 marks]
Total Marks: 7