Question

If the chance that a bus arrives safely at a bus stand is $ \frac{9}{10} $, then the probability that out of 5 buses at least 4 will arrive safely is

• A. \( \frac{12 \times 9^4}{10^5} \)
• B. \( 14 \times \left(\frac{9}{10}\right)^4 \)
• C. \( \frac{14 \times 9^4}{10^5} \)
• D. \( \frac{13 \times 9^4}{10^5} \)

Hint:

For "at least" probability problems in binomial distribution, sum the probabilities of the specified number of successes and all higher numbers of successes up to \( n \). Remember the binomial probability formula: \( P(X=k) = \binom{n}{k} p^k q^{n-k} \).

Answer 1

The Correct Option is C

Step 1: Define the probability of success and failure.
Let \( p \) be the probability that a bus arrives safely.
Given \( p = \frac{9}{10} \).
Let \( q \) be the probability that a bus does not arrive safely.
Then \( q = 1 - p = 1 - \frac{9}{10} = \frac{1}{10} \).
Step 2: Identify the type of probability distribution.
This is a binomial probability problem since there are a fixed number of trials (5 buses), each trial has two possible outcomes (safe arrival or not safe arrival), and the probability of success is constant for each trial.
The binomial probability formula is \( P(X=k) = \binom{n}{k} p^k q^{n-k} \), where \( n \) is the number of trials, and \( k \) is the number of successes.
Step 3: Calculate the probability of at least 4 buses arriving safely.
"At least 4 buses arrive safely" means either exactly 4 buses arrive safely or exactly 5 buses arrive safely.
So, we need to calculate \( P(X \ge 4) = P(X=4) + P(X=5) \). Here, \( n=5 \). For \( P(X=4) \): \[ P(X=4) = \binom{5}{4} p^4 q^{5-4} = \binom{5}{4} p^4 q^1 \] \[ P(X=4) = 5 \times \left(\frac{9}{10}\right)^4 \times \left(\frac{1}{10}\right)^1 = 5 \times \frac{9^4}{10^4} \times \frac{1}{10} = \frac{5 \times 9^4}{10^5} \] For \( P(X=5) \): \[ P(X=5) = \binom{5}{5} p^5 q^{5-5} = \binom{5}{5} p^5 q^0 \] \[ P(X=5) = 1 \times \left(\frac{9}{10}\right)^5 \times (1) = \frac{9^5}{10^5} \] Now, add these probabilities: \[ P(X \ge 4) = P(X=4) + P(X=5) = \frac{5 \times 9^4}{10^5} + \frac{9^5}{10^5} \] Factor out common terms, \( \frac{9^4}{10^5} \): \[ P(X \ge 4) = \frac{9^4}{10^5} (5 + 9) \] \[ P(X \ge 4) = \frac{14 \times 9^4}{10^5} \] The final answer is $\boxed{\text{3}}$.