Question

If 10 percent of the balls in a certain box are red, and if 20 balls are selected from the box at random, with replacement, what is the probability that more than three red balls will be obtained?

Short answer

The probability that more than 3 red balls are obtained is 0.133.

Step-by-step solution

Given information

The number of balls that are selected from the box with replacement is \(n = 20\). The probability that a randomly selected ball in the box is red is \(p = 0.10\).

The balls are collected with replacement.

Compute the probability

Let X be the random variable representing the number of red balls in 20 draws.

In the given scenario, the random variable X will follow the binomial distribution as the trials are independent with fixed trials.

The probability function of a binomial distribution is given as,

\(f\left( x \right) = \left\{ \begin{array}{l}\left( \begin{array}{l}n\\x\end{array} \right){p^x}{\left( {1 - p} \right)^{n - x}}\;\;for\;x = 0,1,...,n,\\0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;otherwise\end{array} \right.\)

For probability of success p in each trial and n fixed trials.

The probability that more than 3 red balls are obtained is computed as,

\(\begin{aligned}{}P\left( {X > 3} \right)& = 1 - P\left( {X \le 3} \right)\\ &= 1 - \left( {P\left( {X = 0} \right) + P\left( {X = 1} \right) + ... + P\left( {X = 3} \right)} \right)\\ &= 1 - \left( {\left( \begin{aligned}{l}20\\0\end{aligned} \right){{\left( {0.10} \right)}^0}{{\left( {1 - 0.10} \right)}^{20 - 0}} + \left( \begin{aligned}{}20\\1\end{aligned} \right){{\left( {0.10} \right)}^1}{{\left( {1 - 0.10} \right)}^{20 - 1}} + ... + \left( \begin{aligned}{l}20\\3\end{aligned} \right){{\left( {0.10} \right)}^3}{{\left( {1 - 0.10} \right)}^{20 - 3}}} \right)\\ &= 1 - \left( {0.12158 + 0.27017 + ... + 0.19012} \right)\\ \approx 0.133\end{aligned}\)

Therefore, the probability that more than 3 red balls are obtained is 0.133.