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There are two urn containing (4 red and 5 black balls) and (5 red and 6 black baltherels ). one ball is drawn at random from the first and transfer to the second and thenone ball is drawn from second and transfer to first. after the transfer one ball is drawn from the first urn , what is the chance that it will be the red ball ?
Verified Answer
There are two urn containing (4 red and 5 black balls) and (5 red and ...
Solution:
P(A) = probability that the ball transferred from urn first to second is black:
=712
P(B) = probability that the ball drawn from second urn is black.
Case I: If white ball goes to urn II
P(C)=512x913
Case II: If black ball goes to urn II
P(D)=712x1013
Thus, P(B)=P(C)+P(D)=(
45156)+(70156)=115
156P(AB)= Probability of event A when B has occurred= Probability that ball drawn from II urn is black
⇒ P(A intersects B)
P(B)=(712x1013)(115156) = 70/115 = 14/23
its just a example.
This question is part of UPSC exam. View all JEE courses
Most Upvoted Answer
There are two urn containing (4 red and 5 black balls) and (5 red and ...
Analysis:
Let's analyze the problem step by step.
Step 1: Calculate the probability of drawing a red ball from the first urn.
In the first urn, there are 4 red balls and 5 black balls. Therefore, the total number of balls in the first urn is 4+5=<4+5=9>>9.
The probability of drawing a red ball from the first urn can be calculated as follows:
P(Red 1st Urn) = Number of red balls in the first urn / Total number of balls in the first urn
= 4/9
So, the probability of drawing a red ball from the first urn is 4/9.
Step 2: Calculate the probability of transferring a red ball from the first urn to the second urn.
After drawing a ball from the first urn, there are now 8 balls left in the first urn. Since one ball is transferred from the first urn to the second urn, the number of red balls and black balls in the first urn will be reduced by 1 each.
Therefore, the probability of transferring a red ball from the first urn to the second urn can be calculated as follows:
P(Red Transfer) = Number of red balls left in the first urn after the transfer / Total number of balls left in the first urn after the transfer
= (4-1) / (8-1)
= 3/7
So, the probability of transferring a red ball from the first urn to the second urn is 3/7.
Step 3: Calculate the probability of transferring a red ball from the second urn to the first urn.
In the second urn, there are 5 red balls and 6 black balls. Since one ball is transferred from the second urn to the first urn, the number of red balls and black balls in the second urn will be reduced by 1 each.
Therefore, the probability of transferring a red ball from the second urn to the first urn can be calculated as follows:
P(Red Transfer) = Number of red balls left in the second urn after the transfer / Total number of balls left in the second urn after the transfer
= (5-1) / (11-1)
= 4/10
= 2/5
So, the probability of transferring a red ball from the second urn to the first urn is 2/5.
Step 4: Calculate the probability of drawing a red ball from the first urn after the transfer.
After the transfer, the first urn will contain 3 red balls and 6 black balls. The total number of balls in the first urn will be 3+6=9.
Therefore, the probability of drawing a red ball from the first urn after the transfer can be calculated as follows:
P(Red 1st Urn after Transfer) = Number of red balls in the first urn after the transfer / Total number of balls in the first urn after the transfer
= 3/9
= 1/3
So, the probability of drawing a red ball from the first urn after the transfer is 1/3.
Conclusion:
The chance of drawing a red ball from the first urn after the transfer is 1/3.
Let's analyze the problem step by step.
Step 1: Calculate the probability of drawing a red ball from the first urn.
In the first urn, there are 4 red balls and 5 black balls. Therefore, the total number of balls in the first urn is 4+5=<4+5=9>>9.
The probability of drawing a red ball from the first urn can be calculated as follows:
P(Red 1st Urn) = Number of red balls in the first urn / Total number of balls in the first urn
= 4/9
So, the probability of drawing a red ball from the first urn is 4/9.
Step 2: Calculate the probability of transferring a red ball from the first urn to the second urn.
After drawing a ball from the first urn, there are now 8 balls left in the first urn. Since one ball is transferred from the first urn to the second urn, the number of red balls and black balls in the first urn will be reduced by 1 each.
Therefore, the probability of transferring a red ball from the first urn to the second urn can be calculated as follows:
P(Red Transfer) = Number of red balls left in the first urn after the transfer / Total number of balls left in the first urn after the transfer
= (4-1) / (8-1)
= 3/7
So, the probability of transferring a red ball from the first urn to the second urn is 3/7.
Step 3: Calculate the probability of transferring a red ball from the second urn to the first urn.
In the second urn, there are 5 red balls and 6 black balls. Since one ball is transferred from the second urn to the first urn, the number of red balls and black balls in the second urn will be reduced by 1 each.
Therefore, the probability of transferring a red ball from the second urn to the first urn can be calculated as follows:
P(Red Transfer) = Number of red balls left in the second urn after the transfer / Total number of balls left in the second urn after the transfer
= (5-1) / (11-1)
= 4/10
= 2/5
So, the probability of transferring a red ball from the second urn to the first urn is 2/5.
Step 4: Calculate the probability of drawing a red ball from the first urn after the transfer.
After the transfer, the first urn will contain 3 red balls and 6 black balls. The total number of balls in the first urn will be 3+6=9.
Therefore, the probability of drawing a red ball from the first urn after the transfer can be calculated as follows:
P(Red 1st Urn after Transfer) = Number of red balls in the first urn after the transfer / Total number of balls in the first urn after the transfer
= 3/9
= 1/3
So, the probability of drawing a red ball from the first urn after the transfer is 1/3.
Conclusion:
The chance of drawing a red ball from the first urn after the transfer is 1/3.
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There are two urn containing (4 red and 5 black balls) and (5 red and 6 black baltherels ). one ball is drawn at random from the first and transfer to the second and thenone ball is drawn from second and transfer to first. after the transfer one ball is drawn from the first urn , what is the chance that it will be the red ball ?
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There are two urn containing (4 red and 5 black balls) and (5 red and 6 black baltherels ). one ball is drawn at random from the first and transfer to the second and thenone ball is drawn from second and transfer to first. after the transfer one ball is drawn from the first urn , what is the chance that it will be the red ball ? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about There are two urn containing (4 red and 5 black balls) and (5 red and 6 black baltherels ). one ball is drawn at random from the first and transfer to the second and thenone ball is drawn from second and transfer to first. after the transfer one ball is drawn from the first urn , what is the chance that it will be the red ball ? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for There are two urn containing (4 red and 5 black balls) and (5 red and 6 black baltherels ). one ball is drawn at random from the first and transfer to the second and thenone ball is drawn from second and transfer to first. after the transfer one ball is drawn from the first urn , what is the chance that it will be the red ball ?.
There are two urn containing (4 red and 5 black balls) and (5 red and 6 black baltherels ). one ball is drawn at random from the first and transfer to the second and thenone ball is drawn from second and transfer to first. after the transfer one ball is drawn from the first urn , what is the chance that it will be the red ball ? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about There are two urn containing (4 red and 5 black balls) and (5 red and 6 black baltherels ). one ball is drawn at random from the first and transfer to the second and thenone ball is drawn from second and transfer to first. after the transfer one ball is drawn from the first urn , what is the chance that it will be the red ball ? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for There are two urn containing (4 red and 5 black balls) and (5 red and 6 black baltherels ). one ball is drawn at random from the first and transfer to the second and thenone ball is drawn from second and transfer to first. after the transfer one ball is drawn from the first urn , what is the chance that it will be the red ball ?.
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There are two urn containing (4 red and 5 black balls) and (5 red and 6 black baltherels ). one ball is drawn at random from the first and transfer to the second and thenone ball is drawn from second and transfer to first. after the transfer one ball is drawn from the first urn , what is the chance that it will be the red ball ?, a detailed solution for There are two urn containing (4 red and 5 black balls) and (5 red and 6 black baltherels ). one ball is drawn at random from the first and transfer to the second and thenone ball is drawn from second and transfer to first. after the transfer one ball is drawn from the first urn , what is the chance that it will be the red ball ? has been provided alongside types of There are two urn containing (4 red and 5 black balls) and (5 red and 6 black baltherels ). one ball is drawn at random from the first and transfer to the second and thenone ball is drawn from second and transfer to first. after the transfer one ball is drawn from the first urn , what is the chance that it will be the red ball ? theory, EduRev gives you an
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