Question

There are two identical urns containing respectively 6 black and 4 red balls, 2 black and 2 red balls. The probability of selecting the first urn is 40%. An urn is chosen at random and a ball is drawn from it find the probability that it is from the first urn given that the ball is black.

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

There are two identical urns containing respectively 6 black and 4 red balls, 2 black and 2 red balls. The probability of selecting the first urn is 40%. An urn is chosen at random and a black ball is drawn from it.

Urn 1 contains 6 black and 4 red balls.

Urn 2 contains 2 black and 2 red balls

Let assume that

E₁ : Urn 1 is selected

E₂ : Urn 2 is selected

E : getting black ball from the bag

Now,

$\rm \: P(E_1) = \dfrac{40}{100} = \dfrac{2}{5}$

$\rm \: P(E_2) = \dfrac{60}{100} = \dfrac{3}{5}$

Now,

$\rm \: P(E |E_1) \: = \: \dfrac{6}{10} = \dfrac{3}{5}$

and

$\rm \: P(E |E_2) \: = \: \dfrac{2}{4} = \dfrac{1}{2}$

So, the required probability of getting black ball from Urn 1 is evaluated by using Baye's Theorem.

So, by using Baye's Theorem, we have

$\rm \: P(E_1 |E) \: = \: \dfrac{P(E_1) . P(E | E_1)}{P(E_1) . P(E | E_1) + P(E_2) . P(E | E_2)}$

So, on substituting the values, we get

$\rm \: =  \: \dfrac{\dfrac{1}{2} \times \dfrac{3}{5} }{\dfrac{1}{2} \times \dfrac{3}{5} + \dfrac{1}{2} \times \dfrac{1}{2} }$

can be further rewritten as

$\rm \: =  \: \dfrac{ \dfrac{3}{5} }{ \dfrac{3}{5} + \dfrac{1}{2} }$

$\rm \: =  \: \dfrac{ \dfrac{3}{5} }{ \dfrac{6 + 5}{10}}$

$\rm \: =  \: \dfrac{3}{5} \times \dfrac{10}{11}$

$\rm \: =  \: \dfrac{6}{11}$

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Formula Used :-

Baye's Theorem states that if S is a sample space and

$\sf \: E_1, \: E_2, \: E_3, \: - - - ,E_n \: are \: mutually \:exclusive \: and \:$

exhaustive events associated with the random experiment and A is any event to occur then

$\\ \: \boxed{\sf{  \: \: \rm \: P(E_i |A) = \dfrac{P(E_i)P(A |E_i)}{\displaystyle\sum_{i=1}^n\rm P(E_i)P(A |E_i)} \: \: }}$

$\rm \: where \: i \: = \: 1, \: 2, \: 3, \: - - - - , \: n$

Answer 2

Step-by-step explanation:

$\color{blue}\begin{array}{|c|c|c|c|} \hline & \colorbox{brown}{\text{ Black Balls }}& \colorbox{brown}{\text{ Red Balls }}& \colorbox{brown}{ \tt \: Total} \\ \hline \colorbox{violet} {\text{ Urn I}} & 6 & 4 & 10 \\ \hline \colorbox{violet}{ \text{ Urn II} }& 2 & 2 & 4 \\ \hline \end{array}$

Let B be the event of getting a black ball.

$\pmb{ \color{darkgreen}\begin{aligned} \text { Now } \quad \tt P(B) \text { from (I) } &=\frac{6}{10} \quad \tt P(I)=P(I I)=\frac{1}{2} \\ \tt P(B) \text { from (II) } & \tt=\frac{2}{4} \\ \tt \therefore \quad P(B) & \tt=P(B)(I) P(I)+P(B)(I I) P(I I) \end{aligned} }$

$\color{darkcyan} \pmb{ \begin{array}{l} \tt=\dfrac{6}{10} \times \dfrac{1}{2}+\dfrac{2}{4} \times \dfrac{1}{2} \\ \\ \tt=\dfrac{3}{10}+ \dfrac{1}{4} \\ \\ \tt = \dfrac{6 + 5}{20} \\ \\ \tt = \dfrac{11}{20} \end{array}}$