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Two type of balls in a bag

Double sampling distributionProbability Math Question…?Probability that the bag contains all balls white given that two balls are whiteConditional Probability about balls in a bagProbability of selecting balls from Bag BProbability question on random selection of balls from bagsWhat is the probability of selecting a white ball given a bag with $3$ red and $5$ white balls and a bag with $4$ and $8$ white balls?Probability of drawing 5 white and 4 black balls and another bag of 7 white and 9 black balls.A bag contains $N$ white balls and $M$ black balls, $3$ balls are chosen at random what is the probabilityA bag contains $6$ white balls and $8$ blue balls.

5

0

$begingroup$

In a bag there are $15$ red and $5$ white balls. Two balls are chosen at random and one is found to be red. The probability that the second one is also red is?

I have attempted this question by counting all the favorable cases:

1. Both red $(15×14)$


2. One red one white $(15×5)$
   

Our case is both red. The probability is, by Baye's theorem, $dfrac15×1415×14+15×5$. However, the answer is not $dfrac1419$ but $dfrac712$.

probability

share|cite|improve this question

edited Nov 24 '18 at 4:36

Saad

20.2k92352

asked Nov 24 '18 at 3:56

CaptainQuestionCaptainQuestion

3089

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well done for sharing an attempt :)
  $endgroup$
  – Shaun
  Nov 24 '18 at 4:07
  

  
  
  
  

add a comment | 

5

0

$begingroup$

In a bag there are $15$ red and $5$ white balls. Two balls are chosen at random and one is found to be red. The probability that the second one is also red is?

I have attempted this question by counting all the favorable cases:

1. Both red $(15×14)$


2. One red one white $(15×5)$
   

Our case is both red. The probability is, by Baye's theorem, $dfrac15×1415×14+15×5$. However, the answer is not $dfrac1419$ but $dfrac712$.

probability

share|cite|improve this question

edited Nov 24 '18 at 4:36

Saad

20.2k92352

asked Nov 24 '18 at 3:56

CaptainQuestionCaptainQuestion

3089

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well done for sharing an attempt :)
  $endgroup$
  – Shaun
  Nov 24 '18 at 4:07
  

  
  
  
  

add a comment | 

$begingroup$

In a bag there are $15$ red and $5$ white balls. Two balls are chosen at random and one is found to be red. The probability that the second one is also red is?

I have attempted this question by counting all the favorable cases:

1. Both red $(15×14)$


2. One red one white $(15×5)$
   

Our case is both red. The probability is, by Baye's theorem, $dfrac15×1415×14+15×5$. However, the answer is not $dfrac1419$ but $dfrac712$.

probability

share|cite|improve this question

edited Nov 24 '18 at 4:36

Saad

20.2k92352

asked Nov 24 '18 at 3:56

CaptainQuestionCaptainQuestion

3089

$endgroup$

share|cite|improve this question

edited Nov 24 '18 at 4:36

Saad

20.2k92352

asked Nov 24 '18 at 3:56

CaptainQuestionCaptainQuestion

3089

share|cite|improve this question

edited Nov 24 '18 at 4:36

Saad

20.2k92352

asked Nov 24 '18 at 3:56

CaptainQuestionCaptainQuestion

3089

edited Nov 24 '18 at 4:36

Saad

20.2k92352

edited Nov 24 '18 at 4:36

Saad

20.2k92352

asked Nov 24 '18 at 3:56

CaptainQuestionCaptainQuestion

3089

asked Nov 24 '18 at 3:56

CaptainQuestionCaptainQuestion

3089

2 Answers
2

active

oldest

votes

2

$begingroup$

You need to include the cases that you choose one white one red, which is the same as $(2)$. Thus, the probability is $$frac15cdot1415cdot14+2cdot15cdot5=frac712.$$

---

Edit: This is assuming that both balls were inspected and (at least) one is found to be red.

The answer is $frac1419$ if one ball was picked from the two chosen and it is found to be red.

share|cite|improve this answer

edited Mar 16 at 22:46

answered Nov 24 '18 at 4:14

Abraham ZhangAbraham Zhang

625412

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, the question is ambiguous as stated, which often happens in these sorts of conditional probability problems. Most of us would assume the condition is "at least one red ball was picked" but it certainly could be "a randomly selected ball from the two picked turned out to be red."
  $endgroup$
  – Ned
  Mar 17 at 1:09
  

  
  
  
  

add a comment | 

1

$begingroup$

In the second case (one red and one white) the count is not correct, since it does not account for the order of the balls. It should be first red and second white, plus first white and second red, in total $15cdot 5 + 5cdot 15 = 150$. So we have $15cdot 4 + 15cdot 5 + 5cdot 15 = 360$ pairs with at least one red ball, and $15*14 = 210$ pairs in which bot are red, hence $P = 210/360 = 7/12$.

share|cite|improve this answer

answered Nov 24 '18 at 4:14

mlerma54mlerma54

1,177148

$endgroup$

add a comment | 

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2

$begingroup$

You need to include the cases that you choose one white one red, which is the same as $(2)$. Thus, the probability is $$frac15cdot1415cdot14+2cdot15cdot5=frac712.$$

---

Edit: This is assuming that both balls were inspected and (at least) one is found to be red.

The answer is $frac1419$ if one ball was picked from the two chosen and it is found to be red.

share|cite|improve this answer

edited Mar 16 at 22:46

answered Nov 24 '18 at 4:14

Abraham ZhangAbraham Zhang

625412

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, the question is ambiguous as stated, which often happens in these sorts of conditional probability problems. Most of us would assume the condition is "at least one red ball was picked" but it certainly could be "a randomly selected ball from the two picked turned out to be red."
  $endgroup$
  – Ned
  Mar 17 at 1:09
  

  
  
  
  

add a comment | 

2

$begingroup$

You need to include the cases that you choose one white one red, which is the same as $(2)$. Thus, the probability is $$frac15cdot1415cdot14+2cdot15cdot5=frac712.$$

---

Edit: This is assuming that both balls were inspected and (at least) one is found to be red.

The answer is $frac1419$ if one ball was picked from the two chosen and it is found to be red.

share|cite|improve this answer

edited Mar 16 at 22:46

answered Nov 24 '18 at 4:14

Abraham ZhangAbraham Zhang

625412

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, the question is ambiguous as stated, which often happens in these sorts of conditional probability problems. Most of us would assume the condition is "at least one red ball was picked" but it certainly could be "a randomly selected ball from the two picked turned out to be red."
  $endgroup$
  – Ned
  Mar 17 at 1:09
  

  
  
  
  

add a comment | 

$begingroup$

You need to include the cases that you choose one white one red, which is the same as $(2)$. Thus, the probability is $$frac15cdot1415cdot14+2cdot15cdot5=frac712.$$

---

Edit: This is assuming that both balls were inspected and (at least) one is found to be red.

The answer is $frac1419$ if one ball was picked from the two chosen and it is found to be red.

share|cite|improve this answer

edited Mar 16 at 22:46

answered Nov 24 '18 at 4:14

Abraham ZhangAbraham Zhang

625412

$endgroup$

share|cite|improve this answer

edited Mar 16 at 22:46

answered Nov 24 '18 at 4:14

Abraham ZhangAbraham Zhang

625412

answered Nov 24 '18 at 4:14

Abraham ZhangAbraham Zhang

625412

answered Nov 24 '18 at 4:14

Abraham ZhangAbraham Zhang

625412

1

$begingroup$

In the second case (one red and one white) the count is not correct, since it does not account for the order of the balls. It should be first red and second white, plus first white and second red, in total $15cdot 5 + 5cdot 15 = 150$. So we have $15cdot 4 + 15cdot 5 + 5cdot 15 = 360$ pairs with at least one red ball, and $15*14 = 210$ pairs in which bot are red, hence $P = 210/360 = 7/12$.

share|cite|improve this answer

answered Nov 24 '18 at 4:14

mlerma54mlerma54

1,177148

$endgroup$

add a comment | 

1

$begingroup$

In the second case (one red and one white) the count is not correct, since it does not account for the order of the balls. It should be first red and second white, plus first white and second red, in total $15cdot 5 + 5cdot 15 = 150$. So we have $15cdot 4 + 15cdot 5 + 5cdot 15 = 360$ pairs with at least one red ball, and $15*14 = 210$ pairs in which bot are red, hence $P = 210/360 = 7/12$.

share|cite|improve this answer

answered Nov 24 '18 at 4:14

mlerma54mlerma54

1,177148

$endgroup$

add a comment | 

$begingroup$

In the second case (one red and one white) the count is not correct, since it does not account for the order of the balls. It should be first red and second white, plus first white and second red, in total $15cdot 5 + 5cdot 15 = 150$. So we have $15cdot 4 + 15cdot 5 + 5cdot 15 = 360$ pairs with at least one red ball, and $15*14 = 210$ pairs in which bot are red, hence $P = 210/360 = 7/12$.

share|cite|improve this answer

answered Nov 24 '18 at 4:14

mlerma54mlerma54

1,177148

$endgroup$

share|cite|improve this answer

answered Nov 24 '18 at 4:14

mlerma54mlerma54

1,177148

answered Nov 24 '18 at 4:14

mlerma54mlerma54

1,177148

answered Nov 24 '18 at 4:14

mlerma54mlerma54

1,177148