Question

Police estimate that 80% of drivers wear their seatbelts. They set up a safety road block stopping, cars to check for seatbelt use. If they stop 9 cars, what is the probability that:
1.1) Exactly three drivers are not wearing their seatbelts?
1.2) At least 3 drivers are not wearing their seatbelt?
1.3) At least 2 drivers but no more than 4 drivers are not wearing their seatbelt?

Answer 1

Answer on Question #51454 – Math – Statistics and Probability

Question

Police estimate that 80% of drivers wear their seatbelts. They set up a safety road block stopping, cars to check for seatbelt use. If they stop 9 cars, what is the probability that:

1.1) Exactly three drivers are not wearing their seatbelts?

1.2) At least 3 drivers are not wearing their seatbelt?

1.3) At least 2 drivers but no more than 4 drivers are not wearing their seatbelt?

Solution

If police estimate that 80% of drivers wear their seatbelts, then we can consider that 20% of drivers don't wear their seatbelts. Then probability:

$$P\left("driver is not wearing seatbelt"\right) = 0.2$$

Let $X$ be a number of drivers who are not wearing seatbelt. Then $X$ - the number of successes in 9 independent experiments. Then using binomial distribution,

$$P(X = k) = C_{9}^{k} p^{k} (1 - p)^{9 - k}$$

where $p = 0.2$, $C_{9}^{k} = \frac{9!}{k! (9 - k)!}$

1.1)

$$P(X = 3) = C_{9}^{3} p^{3} (1 - p)^{6} = \frac{9!}{3!6!} \cdot 0.2^{3} \cdot 0.8^{6} = \frac{7 \cdot 8 \cdot 9}{6} \cdot 0.008 \cdot 0.262144 = 0.176160768$$

1.2) "At least 3 drivers are not wearing their seatbelt" = "3 or 4, or 5, ... or 9 drivers are not wearing their seatbelt"

$$\begin{array}{l} P(X = 3) + P(X = 4) + \dots + P(X = 9) = 1 - P(X = 0) - P(X = 1) - P(X = 2) = \\ = 1 - C_{9}^{0} p^{0} (1 - p)^{9} - C_{9}^{1} p^{1} (1 - p)^{8} - C_{9}^{2} p^{2} (1 - p)^{7} = 1 - 0.8^{9} - 9 \cdot 0.2 \cdot 0.8^{8} - \\ - \frac{8 \cdot 9}{2} \cdot 0.2^{2} \cdot 0.8^{7} = 1 - 0.134217728 - 0.301989888 - 0.301989888 = 0.261800224 \\ \end{array}$$

1.3) "At least 2 drivers but no more than 4 drivers are not wearing their seatbelt?" = "2 or 3, or 4 drivers are not wearing their seatbelt"

$$\begin{array}{l} P (X = 2) + P (X = 3) + P (X = 4) = C _ {0} ^ {2} p ^ {2} (1 - p) ^ {7} + C _ {0} ^ {3} p ^ {3} (1 - p) ^ {6} + C _ {0} ^ {4} p ^ {4} (1 - p) ^ {5} = \\ = p ^ {2} (1 - p) ^ {5} \left(C _ {0} ^ {2} (1 - p) ^ {2} + C _ {0} ^ {3} p (1 - p) + C _ {0} ^ {4} p ^ {2}\right) = 0. 2 ^ {2} 0. 8 ^ {5} (2 3. 0 4 + 1 3. 4 4 + 5. 0 4) = \\ = 0. 5 4 4 2 1 0 9 4 4 \end{array}$$

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