Deepak Scored 45->99%ile with Bounce Back Crack Course. You can do it too!

# An urn contains 25 balls of which 10 balls bear a mark ‘X’ and the remaining 15 bear a mark ‘Y’.

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 is noted down and it is replaced. If 6 balls are drawn in this way, find the probability that

(i) all will bear ‘X’ mark.

(ii) not more than 2 will bear ‘Y’ mark.

(iii) at least one ball will bear ‘Y’ mark

(iv) the number of balls with ‘X’ mark and ‘Y’ mark will be equal.

Solution:

Total number of balls in the urn = 25

Balls bearing mark ‘X’ = 10

Balls bearing mark ‘Y’ = 15

$p=P\left(\right.$ ball bearing mark $\left.' X^{\prime}\right)=\frac{10}{25}=\frac{2}{5}$

q = P (ball bearing mark ‘Y’)$=\frac{15}{25}=\frac{3}{5}$

Six balls are drawn with replacement. Therefore, the number of trials are Bernoulli trials.

Let Z be the random variable that represents the number of balls with ‘Y’ mark on them in the trials.

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

$\therefore \mathrm{P}(\mathrm{Z}=\mathrm{z})={ }^{n} \mathrm{C}_{z} p^{n-z} q^{z}$

(i) $\mathrm{P}$ (all will bear ' $\mathrm{X}$ ' mark) $=\mathrm{P}(\mathrm{Z}=0)={ }^{6} \mathrm{C}_{0}\left(\frac{2}{5}\right)^{6}=\left(\frac{2}{5}\right)^{6}$

(ii) $P$ (not more than 2 bear ' $Y$ ' mark) $=P(Z \leq 2)$

= P (Z = 0) + P (Z = 1) + P (Z = 2)

$={ }^{6} C_{0}(p)^{6}(q)^{0}+{ }^{6} C_{1}(p)^{5}(q)^{1}+{ }^{6} C_{2}(p)^{4}(q)^{2}$

$=\left(\frac{2}{5}\right)^{6}+6\left(\frac{2}{5}\right)^{5}\left(\frac{3}{5}\right)+15\left(\frac{2}{5}\right)^{4}\left(\frac{3}{5}\right)^{2}$

$=\left(\frac{2}{5}\right)^{4}\left[\left(\frac{2}{5}\right)^{2}+6\left(\frac{2}{5}\right)\left(\frac{3}{5}\right)+15\left(\frac{3}{5}\right)^{2}\right]$

$=\left(\frac{2}{5}\right)^{4}\left[\frac{4}{25}+\frac{36}{25}+\frac{135}{25}\right]$

$=\left(\frac{2}{5}\right)^{4}\left[\frac{175}{25}\right]$

$=7\left(\frac{2}{5}\right)^{4}$

(iii) P (at least one ball bears ‘Y’ mark) = P (Z ≥ 1) = 1 − P (Z = 0)

$=1-\left(\frac{2}{5}\right)^{6}$

(iv) P (equal number of balls with ‘X’ mark and ‘Y’ mark) = P (Z = 3)

$={ }^{6} \mathrm{C}_{3}\left(\frac{2}{5}\right)^{3}\left(\frac{3}{5}\right)^{3}$

$=\frac{20 \times 8 \times 27}{15625}$

$=\frac{864}{3125}$