Binomial distribution: Pigs Can Fly

by Kurious Fox

In a magical kingdom, pigs are trained at a prestigious School of Flying with Brooms. However, not all pigs successfully master the art of flying with a broom, and only a certain percentage of them are able to soar through the skies after training.

Suppose that after the training, each pig has a 15% probability (p = 0.15 ) of successfully flying with a broom. You own a mystical farm with 30 magical pigs, all of whom have been trained to fly.

Questions:

  1. What is the probability that exactly 5 pigs out of 30 on your farm can fly with a broom?
  2. What is the probability that at least 1 pig can fly with a broom?
  3. What is the expected number of pigs that can fly with a broom?

Solution:

1. Probability that exactly 5 pigs can fly with a broom:

This is again a binomial probability problem where:

  • n = 30 (total number of pigs),
  • k = 5 (the number of flying pigs we are interested in),
  • p = 0.15 (the probability that a pig can fly with a broom),
  • 1 - p = 0.85 (the probability that a pig cannot fly with a broom).

The formula for binomial probability is:

P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}

Substitute the values:

P(X = 5) = \binom{30}{5} (0.15)^5 (0.85)^{25}

First, calculate the binomial coefficient \binom{30}{5} :

\binom{30}{5} = \frac{30 \times 29 \times 28 \times 27 \times 26}{5 \times 4 \times 3 \times 2 \times 1} = 142506

Now, calculate (0.15)^5 and (0.85)^{25} :

(0.15)^5 \approx 7.59375 \times 10^{-5}, \quad (0.85)^{25} \approx 0.0713

Substitute back into the binomial probability formula:

P(X = 5) \approx 142506 \times 7.59375 \times 10^{-5} \times 0.0713 \approx 0.768

So, the probability that exactly 5 pigs can fly with a broom is approximately 0.0768 or 7.68%.

2. Probability that at least 1 pig can fly with a broom:

The probability that at least one pig can fly is the complement of the probability that none of the pigs can fly. So we calculate:

P(\text{at least 1 can fly}) = 1 - P(\text{none can fly})

The probability that none of the pigs can fly is:

P(\text{none can fly}) = (1 - p)^n = (0.85)^{30}

Now, calculate (0.85)^{30} :

(0.85)^{30} \approx 0.046

Thus, the probability that at least one pig can fly with a broom is:

P(\text{at least 1 can fly}) = 1 - 0.046 = 0.954

So, the probability that at least one pig can fly with a broom is approximately 0.954 or 95.4%.

3. Expected number of pigs that can fly with a broom:

The expected number of pigs that can fly is given by:

E(X) = n \times p

Substitute the values:

E(X) = 30 \times 0.15 = 4.5

So, the expected number of pigs that can fly with a broom is 4.5.

Summary of Answers:

  1. The probability that exactly 5 pigs can fly with a broom is approximately 7.68%.
  2. The probability that at least 1 pig can fly with a broom is approximately 95.4%.
  3. The expected number of pigs that can fly with a broom is 4.5.

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