Question

hw 9 - binomial distributions
score: 4.28/10 answered: 5/10
question 8
for commercial flights in 2010, approximately 10% of flights are late. assuming this success rate
if you randomly select 9 flights, what is the probability that
a) at least one of the flights is late? (round your answer to 4 decimal places)
b) at least two of the flights are late? (round your answer to 4 decimal places)
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Explanation:

Part A

Step1: Identify binomial parameters

We have a binomial distribution with \( n = 9 \) (number of trials), \( p = 0.1 \) (probability of success, i.e., a flight being late), and \( q = 1 - p = 0.9 \) (probability of failure). We want \( P(X \geq 1) \). Using the complement rule, \( P(X \geq 1)=1 - P(X = 0) \).

Step2: Calculate \( P(X = 0) \)

The binomial probability formula is \( P(X = k)=\binom{n}{k}p^{k}q^{n - k} \). For \( k = 0 \), \( \binom{9}{0}=1 \), \( p^{0}=1 \), \( q^{9}=0.9^{9} \). So \( P(X = 0)=1\times1\times0.9^{9} \). Calculate \( 0.9^{9}\approx0.387420489 \).

Step3: Calculate \( P(X \geq 1) \)

\( P(X \geq 1)=1 - 0.387420489\approx0.612579511 \), rounded to 4 decimal places is \( 0.6126 \).

Step1: Use complement rule

We want \( P(X \geq 2) \). Using the complement rule, \( P(X \geq 2)=1 - P(X = 0)-P(X = 1) \). We already know \( P(X = 0)\approx0.387420489 \) from Part A.

Step2: Calculate \( P(X = 1) \)

Using the binomial formula, \( \binom{9}{1}=\frac{9!}{1!(9 - 1)!}=9 \), \( p^{1}=0.1 \), \( q^{8}=0.9^{8} \). So \( P(X = 1)=9\times0.1\times0.9^{8} \). Calculate \( 0.9^{8}\approx0.43046721 \), then \( 9\times0.1\times0.43046721 = 0.387420489 \).

Step3: Calculate \( P(X \geq 2) \)

\( P(X \geq 2)=1 - 0.387420489 - 0.387420489=1 - 0.774840978\approx0.225159022 \), rounded to 4 decimal places is \( 0.2252 \).

Answer:

\( 0.6126 \)

Part B