# An urn contains 10 white and 3 black balls while another urn contains 3 white and 5 black balls. Two balls are drawn from the first urn and put into the second urn then a ball is drawn from the later. Find the probability that it is a white ball.

Answer

Verified

325.8k+ views

Hint: This question can be solved by considering different possibilities of drawing the balls from the first urn and then placing them in the second urn. Then, adding all the probabilities gives the result.

Complete step-by-step answer:

Here, this problem includes the binomial theorem because we require the selection of balls .

Let us look at some definitions:

Mutually Exclusive Events: A set of events is said to be mutually exclusive, if the happening of one event excludes the happening of the other.

If A and B are mutually exclusive events, then \[\left( A\cap B \right)=0\].

Therefore, the probability of mutually exclusive events is \[P\left( A\cap B \right)=\phi \].

Addition Theorem of Probability:

For three events A, B, C

\[P\left( A\cup B\cup C \right)=P\left( A \right)+P\left( B \right)+P\left( C \right)-P\left( A\cap B \right)-P\left( B\cap C \right)-P\left( A\cap C \right)+P\left( A\cap B\cap C \right)\]

Independent Events: Two events A and B associated with a random experiment are independent, if the probability of occurrence of A is not affected by the probability of occurrence of B.

\[P\left( A\cap B \right)=P\left( A \right)P\left( B \right)\]

Now, in this case there are three mutually exclusive and exhaustive ways in which 2 balls can be transferred from one urn to the other.

Let us first consider that 2 white balls are transferred from the first urn to the second.

As we know that from the probability formula:

\[P\left( A \right)=\dfrac{m}{n}=\dfrac{\text{number of favorable outcomes}}{\text{total number of possible outcomes}}\]

As there are 10 white balls and 3 black balls in the first urn. The probability of drawing 2 white balls is:

\[P\left( A \right)=\dfrac{m}{n}\]

\[P\left( A \right)=\dfrac{{}^{10}{{C}_{2}}}{{}^{13}{{C}_{2}}}\]

Now, the 2 white balls drawn from this urn are kept in the other then it has 5 white and 5 black balls.

Here, m=5 and n=10

\[\begin{align}

& P\left( B \right)=\dfrac{m}{n} \\

& P\left( B \right)=\dfrac{5}{10} \\

\end{align}\]

Now, the probability for this whole event is:

\[P\left( A\cap B \right)=P\left( A \right)P\left( B \right)\]

\[\begin{align}

& P\left( A\cap B \right)=\dfrac{{}^{10}{{C}_{2}}}{{}^{13}{{C}_{2}}}\times \dfrac{5}{10} \\

& P\left( A\cap B \right)=\dfrac{45}{78}\times \dfrac{5}{10} \\

& \therefore P\left( A\cap B \right)=\dfrac{225}{780} \\

\end{align}\]

Let us now consider that 2 black balls are transferred from the first urn to the second.

As there are 10 white balls and 3 black balls in the first urn. The probability of drawing 2 black balls is:

\[P\left( A \right)=\dfrac{m}{n}\]

\[P\left( A \right)=\dfrac{{}^{3}{{C}_{2}}}{{}^{13}{{C}_{2}}}\]

Now, the 2 black balls drawn from this urn are kept in the other then it has 3 white and 7 black balls.

Here, m=3 and n=10

\[\begin{align}

& P\left( B \right)=\dfrac{m}{n} \\

& P\left( B \right)=\dfrac{3}{10} \\

\end{align}\]

Now, the probability for this whole event is:

\[P\left( A\cap B \right)=P\left( A \right)P\left( B \right)\]

\[\begin{align}

& P\left( A\cap B \right)=\dfrac{{}^{3}{{C}_{2}}}{{}^{13}{{C}_{2}}}\times \dfrac{3}{10} \\

& P\left( A\cap B \right)=\dfrac{3}{78}\times \dfrac{3}{10} \\

& \therefore P\left( A\cap B \right)=\dfrac{9}{780} \\

\end{align}\]

Let us now consider that 1 black ball and 1 white ball are transferred from the first urn to the second.

As we know that from the probability formula:

\[P\left( A \right)=\dfrac{m}{n}=\dfrac{\text{number of favorable outcomes}}{\text{total number of possible outcomes}}\]

As there are 10 white balls and 3 black balls in the first urn. The probability of drawing 1 black ball and 1 white ball is:

\[P\left( A \right)=\dfrac{m}{n}\]

\[P\left( A \right)=\dfrac{{}^{3}{{C}_{1}}\times {}^{10}{{C}_{1}}}{{}^{13}{{C}_{2}}}\]

Now, in the second urn it becomes 4 white balls and 6 black balls.

Here, m=4, n=10

\[\begin{align}

& P\left( B \right)=\dfrac{m}{n} \\

& P\left( B \right)=\dfrac{4}{10} \\

\end{align}\]

Now, the probability for this whole event is:

\[P\left( A\cap B \right)=P\left( A \right)P\left( B \right)\]

\[\begin{align}

& P\left( A\cap B \right)=\dfrac{{}^{3}{{C}_{1}}\times {}^{10}{{C}_{1}}}{{}^{13}{{C}_{2}}}\times \dfrac{4}{10} \\

& P\left( A\cap B \right)=\dfrac{30}{78}\times \dfrac{4}{10} \\

& \therefore P\left( A\cap B \right)=\dfrac{120}{780} \\

\end{align}\]

Now, by using the formula below we get,

\[P\left( A\cup B\cup C \right)=P\left( A \right)+P\left( B \right)+P\left( C \right)-P\left( A\cap B \right)-P\left( B\cap C \right)-P\left( A\cap C \right)+P\left( A\cap B\cap C \right)\]

\[\therefore \text{total probability}=\dfrac{225+9+120}{780}=\dfrac{354}{780}\]

Note: It is necessary to consider different possible selections of balls from the first urn and then need to be clear how many balls are placed in the second urn of the respective colour.

As the drawing of the balls from the first urn and then placing them in the second urn are two exhaustive events. So, we consider the formula:

\[P\left( A\cap B \right)=P\left( A \right)P\left( B \right)\]

The three different possibilities of drawing the balls from the first urn that we considered here are mutually exclusive to one another because when we consider the occurrence of the first way then there is no chance of the second and third ways.

\[P\left( A\cap B \right)=\phi \]

Complete step-by-step answer:

Here, this problem includes the binomial theorem because we require the selection of balls .

Let us look at some definitions:

Mutually Exclusive Events: A set of events is said to be mutually exclusive, if the happening of one event excludes the happening of the other.

If A and B are mutually exclusive events, then \[\left( A\cap B \right)=0\].

Therefore, the probability of mutually exclusive events is \[P\left( A\cap B \right)=\phi \].

Addition Theorem of Probability:

For three events A, B, C

\[P\left( A\cup B\cup C \right)=P\left( A \right)+P\left( B \right)+P\left( C \right)-P\left( A\cap B \right)-P\left( B\cap C \right)-P\left( A\cap C \right)+P\left( A\cap B\cap C \right)\]

Independent Events: Two events A and B associated with a random experiment are independent, if the probability of occurrence of A is not affected by the probability of occurrence of B.

\[P\left( A\cap B \right)=P\left( A \right)P\left( B \right)\]

Now, in this case there are three mutually exclusive and exhaustive ways in which 2 balls can be transferred from one urn to the other.

Let us first consider that 2 white balls are transferred from the first urn to the second.

As we know that from the probability formula:

\[P\left( A \right)=\dfrac{m}{n}=\dfrac{\text{number of favorable outcomes}}{\text{total number of possible outcomes}}\]

As there are 10 white balls and 3 black balls in the first urn. The probability of drawing 2 white balls is:

\[P\left( A \right)=\dfrac{m}{n}\]

\[P\left( A \right)=\dfrac{{}^{10}{{C}_{2}}}{{}^{13}{{C}_{2}}}\]

Now, the 2 white balls drawn from this urn are kept in the other then it has 5 white and 5 black balls.

Here, m=5 and n=10

\[\begin{align}

& P\left( B \right)=\dfrac{m}{n} \\

& P\left( B \right)=\dfrac{5}{10} \\

\end{align}\]

Now, the probability for this whole event is:

\[P\left( A\cap B \right)=P\left( A \right)P\left( B \right)\]

\[\begin{align}

& P\left( A\cap B \right)=\dfrac{{}^{10}{{C}_{2}}}{{}^{13}{{C}_{2}}}\times \dfrac{5}{10} \\

& P\left( A\cap B \right)=\dfrac{45}{78}\times \dfrac{5}{10} \\

& \therefore P\left( A\cap B \right)=\dfrac{225}{780} \\

\end{align}\]

Let us now consider that 2 black balls are transferred from the first urn to the second.

As there are 10 white balls and 3 black balls in the first urn. The probability of drawing 2 black balls is:

\[P\left( A \right)=\dfrac{m}{n}\]

\[P\left( A \right)=\dfrac{{}^{3}{{C}_{2}}}{{}^{13}{{C}_{2}}}\]

Now, the 2 black balls drawn from this urn are kept in the other then it has 3 white and 7 black balls.

Here, m=3 and n=10

\[\begin{align}

& P\left( B \right)=\dfrac{m}{n} \\

& P\left( B \right)=\dfrac{3}{10} \\

\end{align}\]

Now, the probability for this whole event is:

\[P\left( A\cap B \right)=P\left( A \right)P\left( B \right)\]

\[\begin{align}

& P\left( A\cap B \right)=\dfrac{{}^{3}{{C}_{2}}}{{}^{13}{{C}_{2}}}\times \dfrac{3}{10} \\

& P\left( A\cap B \right)=\dfrac{3}{78}\times \dfrac{3}{10} \\

& \therefore P\left( A\cap B \right)=\dfrac{9}{780} \\

\end{align}\]

Let us now consider that 1 black ball and 1 white ball are transferred from the first urn to the second.

As we know that from the probability formula:

\[P\left( A \right)=\dfrac{m}{n}=\dfrac{\text{number of favorable outcomes}}{\text{total number of possible outcomes}}\]

As there are 10 white balls and 3 black balls in the first urn. The probability of drawing 1 black ball and 1 white ball is:

\[P\left( A \right)=\dfrac{m}{n}\]

\[P\left( A \right)=\dfrac{{}^{3}{{C}_{1}}\times {}^{10}{{C}_{1}}}{{}^{13}{{C}_{2}}}\]

Now, in the second urn it becomes 4 white balls and 6 black balls.

Here, m=4, n=10

\[\begin{align}

& P\left( B \right)=\dfrac{m}{n} \\

& P\left( B \right)=\dfrac{4}{10} \\

\end{align}\]

Now, the probability for this whole event is:

\[P\left( A\cap B \right)=P\left( A \right)P\left( B \right)\]

\[\begin{align}

& P\left( A\cap B \right)=\dfrac{{}^{3}{{C}_{1}}\times {}^{10}{{C}_{1}}}{{}^{13}{{C}_{2}}}\times \dfrac{4}{10} \\

& P\left( A\cap B \right)=\dfrac{30}{78}\times \dfrac{4}{10} \\

& \therefore P\left( A\cap B \right)=\dfrac{120}{780} \\

\end{align}\]

Now, by using the formula below we get,

\[P\left( A\cup B\cup C \right)=P\left( A \right)+P\left( B \right)+P\left( C \right)-P\left( A\cap B \right)-P\left( B\cap C \right)-P\left( A\cap C \right)+P\left( A\cap B\cap C \right)\]

\[\therefore \text{total probability}=\dfrac{225+9+120}{780}=\dfrac{354}{780}\]

Note: It is necessary to consider different possible selections of balls from the first urn and then need to be clear how many balls are placed in the second urn of the respective colour.

As the drawing of the balls from the first urn and then placing them in the second urn are two exhaustive events. So, we consider the formula:

\[P\left( A\cap B \right)=P\left( A \right)P\left( B \right)\]

The three different possibilities of drawing the balls from the first urn that we considered here are mutually exclusive to one another because when we consider the occurrence of the first way then there is no chance of the second and third ways.

\[P\left( A\cap B \right)=\phi \]

Last updated date: 28th May 2023

â€¢

Total views: 325.8k

â€¢

Views today: 6.84k

Recently Updated Pages

If ab and c are unit vectors then left ab2 right+bc2+ca2 class 12 maths JEE_Main

A rod AB of length 4 units moves horizontally when class 11 maths JEE_Main

Evaluate the value of intlimits0pi cos 3xdx A 0 B 1 class 12 maths JEE_Main

Which of the following is correct 1 nleft S cup T right class 10 maths JEE_Main

What is the area of the triangle with vertices Aleft class 11 maths JEE_Main

KCN reacts readily to give a cyanide with A Ethyl alcohol class 12 chemistry JEE_Main

Trending doubts

What was the capital of Kanishka A Mathura B Purushapura class 7 social studies CBSE

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Tropic of Cancer passes through how many states? Name them.

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

Name the Largest and the Smallest Cell in the Human Body ?