What is the probability that the first white ball is seen after the 6th draw?

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An urn contains $3$ white balls and $7$ red balls. Balls are drawn from the urn one by one and without replacement.What is the probability that the first white ball is seen after the $6$th draw?

My analysis:
The probability of picking the first red ball is :7/10
The probability of picking the second red ball is :6/9
The probability of picking the third red ball is :5/8
The probability of picking the fourth red ball is :4/7
The probability of picking the fifth red ball is :3/6
The probability of picking the first red ball is 2/5
And the probability of picking the white ball after all is 1/4
Multiplying all since the draws are independent gives me:1/120 as an answer whereas the true answer must be: 1/30

probability

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edited Jul 14 at 23:54

Key Flex

4,451525

asked Jul 14 at 23:24

Ri Ray

51

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The last step is wrong. You calculated (incorrectly) the probability that the first white appeared on draw $7$. That isn't what was asked.
  – lulu
  Jul 14 at 23:37
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

An urn contains $3$ white balls and $7$ red balls. Balls are drawn from the urn one by one and without replacement.What is the probability that the first white ball is seen after the $6$th draw?

My analysis:
The probability of picking the first red ball is :7/10
The probability of picking the second red ball is :6/9
The probability of picking the third red ball is :5/8
The probability of picking the fourth red ball is :4/7
The probability of picking the fifth red ball is :3/6
The probability of picking the first red ball is 2/5
And the probability of picking the white ball after all is 1/4
Multiplying all since the draws are independent gives me:1/120 as an answer whereas the true answer must be: 1/30

probability

share|cite|improve this question

edited Jul 14 at 23:54

Key Flex

4,451525

asked Jul 14 at 23:24

Ri Ray

51

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The last step is wrong. You calculated (incorrectly) the probability that the first white appeared on draw $7$. That isn't what was asked.
  – lulu
  Jul 14 at 23:37
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

An urn contains $3$ white balls and $7$ red balls. Balls are drawn from the urn one by one and without replacement.What is the probability that the first white ball is seen after the $6$th draw?

My analysis:
The probability of picking the first red ball is :7/10
The probability of picking the second red ball is :6/9
The probability of picking the third red ball is :5/8
The probability of picking the fourth red ball is :4/7
The probability of picking the fifth red ball is :3/6
The probability of picking the first red ball is 2/5
And the probability of picking the white ball after all is 1/4
Multiplying all since the draws are independent gives me:1/120 as an answer whereas the true answer must be: 1/30

probability

share|cite|improve this question

edited Jul 14 at 23:54

Key Flex

4,451525

asked Jul 14 at 23:24

Ri Ray

51

An urn contains $3$ white balls and $7$ red balls. Balls are drawn from the urn one by one and without replacement.What is the probability that the first white ball is seen after the $6$th draw?

My analysis:
The probability of picking the first red ball is :7/10
The probability of picking the second red ball is :6/9
The probability of picking the third red ball is :5/8
The probability of picking the fourth red ball is :4/7
The probability of picking the fifth red ball is :3/6
The probability of picking the first red ball is 2/5
And the probability of picking the white ball after all is 1/4
Multiplying all since the draws are independent gives me:1/120 as an answer whereas the true answer must be: 1/30

probability

share|cite|improve this question

edited Jul 14 at 23:54

Key Flex

4,451525

asked Jul 14 at 23:24

Ri Ray

51

share|cite|improve this question

share|cite|improve this question

edited Jul 14 at 23:54

Key Flex

4,451525

edited Jul 14 at 23:54

Key Flex

4,451525

edited Jul 14 at 23:54

Key Flex

4,451525

4,451525

asked Jul 14 at 23:24

Ri Ray

51

asked Jul 14 at 23:24

Ri Ray

51

asked Jul 14 at 23:24

Ri Ray

51

51

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The last step is wrong. You calculated (incorrectly) the probability that the first white appeared on draw $7$. That isn't what was asked.
  – lulu
  Jul 14 at 23:37
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The last step is wrong. You calculated (incorrectly) the probability that the first white appeared on draw $7$. That isn't what was asked.
  – lulu
  Jul 14 at 23:37
  

  
  
  
  

1

1

The last step is wrong. You calculated (incorrectly) the probability that the first white appeared on draw $7$. That isn't what was asked.
– lulu
Jul 14 at 23:37

The last step is wrong. You calculated (incorrectly) the probability that the first white appeared on draw $7$. That isn't what was asked.
– lulu
Jul 14 at 23:37

add a comment | 

1 Answer
1

active

oldest

votes

up vote
0
down vote



accepted

Note that in the problem it clearly mentioned that the white ball is seen only after the $6^th$ draw. So basically you just need to find the probability that the first $6$ balls drawn are red.

Edit:

Probability that the first drawn to be red is $dfrac710$

Probability that the second drawn to be red is $dfrac69$

Probability that the third drawn to be red is $dfrac58$

Probability that the fourth drawn to be red is $dfrac47$

Probability that the fifth drawn to be red is $dfrac36$

Probability that the sixth drawn to be red is $dfrac25$

Now the probability is $dfrac710timesdfrac69timesdfrac58timesdfrac47timesdfrac36timesdfrac25=dfrac130$

share|cite|improve this answer

edited Jul 14 at 23:53

answered Jul 14 at 23:31

Key Flex

4,451525

• 
  
  
  

  
  

  
  
  
  
  
  

  
  That’s whay I did but I couldn’t reach the same answer
  – Ri Ray
  Jul 14 at 23:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @RiRay See the edit.
  – Key Flex
  Jul 14 at 23:53
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote



accepted

Note that in the problem it clearly mentioned that the white ball is seen only after the $6^th$ draw. So basically you just need to find the probability that the first $6$ balls drawn are red.

Edit:

Probability that the first drawn to be red is $dfrac710$

Probability that the second drawn to be red is $dfrac69$

Probability that the third drawn to be red is $dfrac58$

Probability that the fourth drawn to be red is $dfrac47$

Probability that the fifth drawn to be red is $dfrac36$

Probability that the sixth drawn to be red is $dfrac25$

Now the probability is $dfrac710timesdfrac69timesdfrac58timesdfrac47timesdfrac36timesdfrac25=dfrac130$

share|cite|improve this answer

edited Jul 14 at 23:53

answered Jul 14 at 23:31

Key Flex

4,451525

• 
  
  
  

  
  

  
  
  
  
  
  

  
  That’s whay I did but I couldn’t reach the same answer
  – Ri Ray
  Jul 14 at 23:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @RiRay See the edit.
  – Key Flex
  Jul 14 at 23:53
  

  
  
  
  

add a comment | 

up vote
0
down vote



accepted

Note that in the problem it clearly mentioned that the white ball is seen only after the $6^th$ draw. So basically you just need to find the probability that the first $6$ balls drawn are red.

Edit:

Probability that the first drawn to be red is $dfrac710$

Probability that the second drawn to be red is $dfrac69$

Probability that the third drawn to be red is $dfrac58$

Probability that the fourth drawn to be red is $dfrac47$

Probability that the fifth drawn to be red is $dfrac36$

Probability that the sixth drawn to be red is $dfrac25$

Now the probability is $dfrac710timesdfrac69timesdfrac58timesdfrac47timesdfrac36timesdfrac25=dfrac130$

share|cite|improve this answer

edited Jul 14 at 23:53

answered Jul 14 at 23:31

Key Flex

4,451525

• 
  
  
  

  
  

  
  
  
  
  
  

  
  That’s whay I did but I couldn’t reach the same answer
  – Ri Ray
  Jul 14 at 23:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @RiRay See the edit.
  – Key Flex
  Jul 14 at 23:53
  

  
  
  
  

add a comment | 

up vote
0
down vote



accepted

up vote
0
down vote



accepted

Note that in the problem it clearly mentioned that the white ball is seen only after the $6^th$ draw. So basically you just need to find the probability that the first $6$ balls drawn are red.

Edit:

Probability that the first drawn to be red is $dfrac710$

Probability that the second drawn to be red is $dfrac69$

Probability that the third drawn to be red is $dfrac58$

Probability that the fourth drawn to be red is $dfrac47$

Probability that the fifth drawn to be red is $dfrac36$

Probability that the sixth drawn to be red is $dfrac25$

Now the probability is $dfrac710timesdfrac69timesdfrac58timesdfrac47timesdfrac36timesdfrac25=dfrac130$

share|cite|improve this answer

edited Jul 14 at 23:53

answered Jul 14 at 23:31

Key Flex

4,451525

Note that in the problem it clearly mentioned that the white ball is seen only after the $6^th$ draw. So basically you just need to find the probability that the first $6$ balls drawn are red.

Edit:

Probability that the first drawn to be red is $dfrac710$

Probability that the second drawn to be red is $dfrac69$

Probability that the third drawn to be red is $dfrac58$

Probability that the fourth drawn to be red is $dfrac47$

Probability that the fifth drawn to be red is $dfrac36$

Probability that the sixth drawn to be red is $dfrac25$

Now the probability is $dfrac710timesdfrac69timesdfrac58timesdfrac47timesdfrac36timesdfrac25=dfrac130$

share|cite|improve this answer

edited Jul 14 at 23:53

answered Jul 14 at 23:31

Key Flex

4,451525

share|cite|improve this answer

share|cite|improve this answer

edited Jul 14 at 23:53

edited Jul 14 at 23:53

edited Jul 14 at 23:53

answered Jul 14 at 23:31

Key Flex

4,451525

answered Jul 14 at 23:31

Key Flex

4,451525

answered Jul 14 at 23:31

Key Flex

4,451525

4,451525

• 
  
  
  

  
  

  
  
  
  
  
  

  
  That’s whay I did but I couldn’t reach the same answer
  – Ri Ray
  Jul 14 at 23:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @RiRay See the edit.
  – Key Flex
  Jul 14 at 23:53
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  That’s whay I did but I couldn’t reach the same answer
  – Ri Ray
  Jul 14 at 23:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @RiRay See the edit.
  – Key Flex
  Jul 14 at 23:53
  

  
  
  
  

That’s whay I did but I couldn’t reach the same answer
– Ri Ray
Jul 14 at 23:35

That’s whay I did but I couldn’t reach the same answer
– Ri Ray
Jul 14 at 23:35

@RiRay See the edit.
– Key Flex
Jul 14 at 23:53

@RiRay See the edit.
– Key Flex
Jul 14 at 23:53

add a comment | 

 

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