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.

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Solution

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, $p = \frac{2}{5} a n d q = \frac{3}{5}$

$∴$ $P (Z = r) =^{n} C_{r} p^{r} q^{n - r} =^{6} C_{r} {(\frac{2}{5})}^{r} {(\frac{3}{5})}^{6 - r}$

Required probability =P(three successes and three failures)

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

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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.

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.