Question

An urn contains $25$ balls of which $10$ bear a mark '$X$' and the remaining $15$ bear a mark '$Y$'. A ball is drawn at random from the urn, its mark noted down and it is replaced. If $6$ balls are drawn in this way, find the probability that
(i) All will bear mark '$X$'
(ii) Not more than two will bear '$Y$' mark

Answer 1

From the given statement :

Total number of balls = $25$

Balls bear mark $Y=15$

Balls bear mark $X=10$

Probability that balls bear mark $X=p=\frac{10}{25}=\frac{2}{5}$

Probability that balls bear mark $Y=q=\frac{15}{25}=\frac{3}{5}$

Now 6 balls are drawn with replacement

So, the number of trials are Bernoulli trials.

Let Z is the random variable that represents the number of balls with X mark on them in the trial.

Hence, Z has the binomial distribution with $n=6,p=\frac{2}{5}$

$P(Z=z){=}^n{\text{c}}_z\times p^{n-z}\times q^z$

1.Probability that all balls bear mark $X=P(Z=0){=}^6{\text{c}}_0\times {\left(\frac{2}{5}\right)}^{6-0}\times {\left(\frac{3}{5}\right)}^0$

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

2. Probabiliy that not more than 2 balls bear mark $Y=P(Z<=2)=P(Z=0)+P(Z=1)+P(Z=2)$

= ${}^6{\text{c}}_0\times {\left(\frac{2}{5}\right)}^{6-0}\times {\left(\frac{3}{5}\right)}^0{+}^6{\text{c}}_1\times {\left(\frac{2}{5}\right)}^{6-1}\times {\left(\frac{3}{5}\right)}^1{+}^6{\text{c}}_2\times {\left(\frac{2}{5}\right)}^{6-2}\times {\left(\frac{3}{5}\right)}^2$

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

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

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