A bag consists of 10 balls each marked with one of the digits 0 to 9.

Question:

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:

Let X denote the number of balls marked with the digit 0 among the 4 balls drawn.

Since the balls are drawn with replacement, the trials are Bernoulli trials.

$X$ has a binomial distribution with $n=4$ and $p=\frac{1}{10}$

$\therefore q=1-p=1-\frac{1}{10}=\frac{9}{10}$

$\therefore \mathrm{P}(\mathrm{X}=x)={ }^{n} \mathrm{C}_{x} q^{v-x} \cdot p^{x}, x=1,2, \ldots n$

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

P (none marked with 0) = P (X = 0)

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

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

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

 

 

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