Question

Every time I use a piece of scrap paper, I crumple it up and try to shoot it inside the recycling bin across the room. I'm pretty good at it: If I shoot $5$ pieces of paper at the recycling bin, at least one of them will make it inside the recycling bin with probability $\frac{211}{243}$. If I shoot $6$ pieces of paper at the recycling bin, what's the probability at least two of them make it inside the recycling bin

Answer 1

Using the binomial distribution, it is found that there is a 0.0012 = 0.12% probability at least two of them make it inside the recycling bin.

What is the binomial distribution formula?

The formula is:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

• x is the number of successes.

• n is the number of trials.

• p is the probability of a success on a single trial.

With 5 shoots, the probability of making at least one is $\frac{211}{243}$, hence the probability of making none, P(X = 0), is $\frac{232}{243}$, hence:

$(1 - p)^5 = \frac{232}{243}$

$\sqrt[5]{(1 - p)^5} = \sqrt[5]{\frac{232}{243}}$

1 - p = 0.9908

p = 0.0092

Then, with 6 shoots, the parameters are:

n = 6, p = 0.0092.

The probability that at least two of them make it inside the recycling bin is:

$P(X \geq 2) = 1 - P(X < 2)$

In which:

[P(X < 2) = P(X = 0) + P(X = 1)

Then:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{6,0}.(0.0092)^{0}.(0.9908)^{6} = 0.9461$

$P(X = 1) = C_{6,1}.(0.0092)^{1}.(0.9908)^{5} = 0.0527$

Then:

P(X < 2) = P(X = 0) + P(X = 1) = 0.9461 + 0.0527 = 0.9988

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.9988 = 0.0012$

0.0012 = 0.12% probability at least two of them make it inside the recycling bin.

More can be learned about the binomial distribution at https://brainly.com/question/24863377

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