Question

An urn contains 25 balls of which 10 balls bear a mark X and the remaining 15 bear a mark Y. A ball is drawn at random from the urn, its mark note down and it is replaced. If 6 balls are drawn in this way, find the probability that

atleast one ball will bear Y mark

Answer 1

It is case of Bernoulli trials with n=6. Let success be defined as drawing a ball marked X.

p=P(a success in a single draw)=$\frac{10}{25}=\frac{2}{5}$

$\Rightarrow q=1-\frac{2}{5}=\frac{3}{5}$

Clearly, Z has a binomial distribution with n=6, $p=\frac{2}{5}andq=\frac{3}{5}$

$\therefore$ $P(Z=r){=}^n{C}_r{p}^r{q}^{n-r}{=}^6{C}_r{\left(\frac{2}{5}\right)}^r{\left(\frac{3}{5}\right)}^{6-r}$

P (atleast on eball will bear mark Y) = P(atleast one failure)

=P(atmost five successes)

$=1-P\left(6\right)=1{-}^6{C}_6{p}^6{q}^0=1-1\times {\left(\frac{2}{5}\right)}^6\times =1-{\left(\frac{2}{5}\right)}^6$