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

all will bear X mark

not more than 2 will bear Y mark

atleast one ball will bear Y mark

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

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

P (all bear mark X)=P(6 success)

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

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(not more than 2 bear mark Y)

=P(not less than 4 bear mark X)= P(atleast 4 successes)

$P\left(4\right)+P\left(5\right)+P\left(6\right){=}^6{C}_4{p}^4{q}^2{+}^6{C}_5{p}^{5}q{+}^6{C}_6{p}^6{q}^0$

${=}^6{C}_4{\left(\frac{2}{5}\right)}^4{\left(\frac{3}{5}\right)}^2{+}^6{C}_5{\left(\frac{2}{5}\right)}^5{\left(\frac{3}{5}\right)}^1{+}^6{C}_6{\left(\frac{2}{5}\right)}^6{\left(\frac{3}{5}\right)}^0=15\times {\left(\frac{2}{5}\right)}^4\frac{9}{25}+6\times {\left(\frac{2}{5}\right)}^5\frac{3}{5}+1\times {\left(\frac{2}{5}\right)}^6\times 1={\left(\frac{2}{5}\right)}^4[\frac{27}{5}+\frac{36}{25}+\frac{4}{25}]={\left(\frac{2}{5}\right)}^4\left[\frac{135+36+4}{25}\right]={\left(\frac{2}{5}\right)}^4\times \frac{175}{25}=7{\left(\frac{2}{5}\right)}^4$

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)

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

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)

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