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

The number of balls with X mark and Y mark will be equal.

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, $p=\frac{2}{5}andq=\frac{3}{5}$

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

Required probability =P(three successes and three failures)

$=P\left(3\right){=}^6{C}_3{p}^3{q}^3{=}^6{C}_3{\left(\frac{2}{5}\right)}^3{\left(\frac{3}{5}\right)}^3=20\times {\left(\frac{2}{5}\right)}^3{\left(\frac{3}{5}\right)}^3=20\times \frac{8}{125}\times \frac{27}{125}=\frac{864}{3125}$