Question

If 50 (fifty) classes of 20 (twenty) students are randomly selected, what is the probability that 10 (ten) classes have no left-handed students if 8% of students in the class are left-handed students.

Answer 1

Answer on Question #69532 – Math – Statistics and Probability

Question

If 50 (fifty) classes of 20 (twenty) students are randomly selected, what is the probability that 10 (ten) classes have no left-handed students if 8% of students in the class are left-handed students.

Solution

Let's recall the formula for binomial probability (probability for Bernoulli experiments):

$$P(k) = \frac{n!}{k! (n - k)!} p^k (1 - p)^{n - k},$$

where

- $P(k)$ is the probability of $k$ successes out of $n$ experiments,

- $n$ is the number of experiments,

- $p$ is the probability of success,

- and $(1 - p)$ is the probability of failure.

1. Let $k = 20$, $n = 20$, $p = 1 - 0.08$. Then the probability that a class has all 20 (twenty) no left-handed students is

$$P_1 = \frac{20!}{20! 0!} (0.92)^{20} (0.08)^0 \approx 0.19.$$

2. Let $k = 10$, $n = 50$, $p = P_1 = 0.19$. Then the probability that 10 (ten) selected classes have no left-handed students is

$$P_2 = \frac{50!}{10! 40!} (0.19)^{10} (0.81)^{40} \approx 0.14.$$

Answer:

The probability is 0.14.

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