A bag consists of 10 balls each marked with one of the digits 0 to 9. If four balls are drawn successively with replacement from the bag, what is the probability that none is marked with the digit 0?
A bag consists of 10 balls each marked with one of the digits 0 to 9. If four balls are drawn successively with replacement from the bag, what is the probability that none is marked with the digit 0?

Solution:

Assume that the number of balls with a digit marked as zero in the experiment of four balls drawn at the same time is x.

The trial is a Bernoulli trial because the balls are drawn with replacement, as can be seen.

Probability of a ball drawn from the bag to be marked as digit 0=1 / 10

It can be clearly observed that \mathrm{X} has a binomial distribution with \mathrm{n}=4 and \mathrm{p}=1 / 10

Thus, q=1-p=1-1 / 10=9 / 10

Thus, \mathrm{P}(\mathrm{X}=\mathrm{x})={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{x}} \mathrm{q}^{\mathrm{n}-\mathrm{x}} \mathrm{p}^{\mathrm{x}}, where \mathrm{x}=0,1,2, \ldots \mathrm{n}

={ }^{4} \mathrm{C}_{\mathrm{x}}\left(\frac{9}{10}\right)^{4-\mathrm{x}}\left(\frac{1}{10}\right)^{\mathrm{x}}

Probability of no ball marked with zero among the 4 balls =\mathrm{P}(\mathrm{X}=0)

={ }^{4} C_{0}\left(\frac{9}{10}\right)^{4-0}\left(\frac{1}{10}\right)^{0}

={ }^{4} \mathrm{C}_{\mathrm{x}}\left(\frac{9}{10}\right)^{4}\left(\frac{1}{10}\right)^{0}

=1 \times\left(\frac{9}{10}\right)^{4}

=\left(\frac{9}{10}\right)^{4}