91Ó°ÊÓ

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.

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.