Probability with Distributions

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A bin has 10 red balls and 8 blue balls. We randomly draw out 6 balls, one at a time, without replacement. What is the probability, that, at some point, we choose two consecutive balls that are red?

This is an interesting distribution problem, and here is how I started. So first I did $binom10+86 = binom186$ which is the number of ways to pick 6 balls out of 18 without regard to order. Now I think I should do casework based on the amount of red balls I draw? Obviously 0 or 1 red balls won't work.

How should I continue?

probability

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edited Jul 24 at 2:27

asked Jul 24 at 2:13

TheLeogend

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up vote
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A bin has 10 red balls and 8 blue balls. We randomly draw out 6 balls, one at a time, without replacement. What is the probability, that, at some point, we choose two consecutive balls that are red?

This is an interesting distribution problem, and here is how I started. So first I did $binom10+86 = binom186$ which is the number of ways to pick 6 balls out of 18 without regard to order. Now I think I should do casework based on the amount of red balls I draw? Obviously 0 or 1 red balls won't work.

How should I continue?

probability

share|cite|improve this question

edited Jul 24 at 2:27

asked Jul 24 at 2:13

TheLeogend

309112

add a comment | 

up vote
-1
down vote

favorite

up vote
-1
down vote

favorite

A bin has 10 red balls and 8 blue balls. We randomly draw out 6 balls, one at a time, without replacement. What is the probability, that, at some point, we choose two consecutive balls that are red?

This is an interesting distribution problem, and here is how I started. So first I did $binom10+86 = binom186$ which is the number of ways to pick 6 balls out of 18 without regard to order. Now I think I should do casework based on the amount of red balls I draw? Obviously 0 or 1 red balls won't work.

How should I continue?

probability

share|cite|improve this question

edited Jul 24 at 2:27

asked Jul 24 at 2:13

TheLeogend

309112

A bin has 10 red balls and 8 blue balls. We randomly draw out 6 balls, one at a time, without replacement. What is the probability, that, at some point, we choose two consecutive balls that are red?

This is an interesting distribution problem, and here is how I started. So first I did $binom10+86 = binom186$ which is the number of ways to pick 6 balls out of 18 without regard to order. Now I think I should do casework based on the amount of red balls I draw? Obviously 0 or 1 red balls won't work.

How should I continue?

probability

share|cite|improve this question

edited Jul 24 at 2:27

asked Jul 24 at 2:13

TheLeogend

309112

share|cite|improve this question

share|cite|improve this question

edited Jul 24 at 2:27

edited Jul 24 at 2:27

edited Jul 24 at 2:27

asked Jul 24 at 2:13

TheLeogend

309112

asked Jul 24 at 2:13

TheLeogend

309112

asked Jul 24 at 2:13

TheLeogend

309112

309112

add a comment | 

add a comment | 

3 Answers
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It may be easier to start by finding the probability that no red balls are adjacent.

There are $binom186 6!$ possible sequences of the selected balls, all of which we assume are equally likely.

Suppose there are $i$ red balls among the $6$ selected. If no red balls are adjacent, then we must have $0 le i le 3$, and the positions of the red balls can be chosen in $binom6-i+1i$ ways. Once the positions of the $i$ red balls are chosen, they can be selected and sequenced in $binom10i i!$ ways, and the $6-i$ blue balls can be selected and sequenced in $binom86-i (6-i)! $ ways. So the total number of sequences in which no red balls are adjacent is
$$N = sum_i=0^3 binom6-i+1i ; binom10i i! ; binom86-i (6-i)!$$
and the probability that no red balls are adjacent is
$$p = fracN binom186 6! = 0.217195$$
So the probability that at least two red balls are adjacent is $1-p = boxed0.782805.
$

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edited Jul 28 at 14:34

answered Jul 24 at 17:16

awkward

5,12111021

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I would attempt this problem using the method of complementary set. Essentially, I try to find the number of ways of drawing 6 balls such that 2 consecutive red balls do not appear, and subtract this probability from 1

Now, if I draw a red ball at a particular point, the next ball I draw must be a blue ball (so that no two consecutive red balls are drawn)

Using casework, we can consider the possibilities for each ball (either red or blue), which is not as difficult as it may seem given that after a red ball there must be a blue ball

Can you continue from here?

share|cite|improve this answer

answered Jul 24 at 2:20

Vikram

1716

add a comment | 

up vote
0
down vote

The six balls that you draw in terms of Red and Blue are

Case 1: R - 6, B - 0: The probability that such a sequence of draw $P_1 = frac10choose68choose018choose6$

Case 2: R - 5, B - 1 : $P_2 = frac10choose58choose118choose6$

Case 3: R - 4, B - 2 : $P_3 = frac10choose48choose218choose6$

Case 4: R - 3, B - 3 : $P_4 = frac10choose38choose318choose6$

Case 5: R - 2, B - 4 : $P_5 = frac10choose28choose418choose6$

Case 6: R - 1, B - 5

Case 7: R - 0, B - 6

In cases 1, 2, 3, there at some point you would have two consecutive reds drawn.

Only in cases 4 and 5 do you have times when you don't have two consecutive reds and the probability of that would be $PP_4= frac4choose36choose3$ and in case 5,$PP_5= frac5choose26choose2$and 1 minus that will be the probability that you will find atleast one time two consecutive red. In case 4, it is $1-frac15$ and in case 5, it is $1-frac1015$. IN the last two cases there are no ways to have two consecutive reds.

Thus the required probability $= P_1+P_2+P_3+P_4times(1-PP_4)+P_5times(1-PP_5).=0.78280543$

share|cite|improve this answer

edited Jul 24 at 7:29

answered Jul 24 at 6:13

Satish Ramanathan

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3 Answers
3

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3 Answers
3

active

oldest

votes

active

oldest

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active

oldest

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up vote
1
down vote

It may be easier to start by finding the probability that no red balls are adjacent.

There are $binom186 6!$ possible sequences of the selected balls, all of which we assume are equally likely.

Suppose there are $i$ red balls among the $6$ selected. If no red balls are adjacent, then we must have $0 le i le 3$, and the positions of the red balls can be chosen in $binom6-i+1i$ ways. Once the positions of the $i$ red balls are chosen, they can be selected and sequenced in $binom10i i!$ ways, and the $6-i$ blue balls can be selected and sequenced in $binom86-i (6-i)! $ ways. So the total number of sequences in which no red balls are adjacent is
$$N = sum_i=0^3 binom6-i+1i ; binom10i i! ; binom86-i (6-i)!$$
and the probability that no red balls are adjacent is
$$p = fracN binom186 6! = 0.217195$$
So the probability that at least two red balls are adjacent is $1-p = boxed0.782805.
$

share|cite|improve this answer

edited Jul 28 at 14:34

answered Jul 24 at 17:16

awkward

5,12111021

add a comment | 

up vote
1
down vote

It may be easier to start by finding the probability that no red balls are adjacent.

There are $binom186 6!$ possible sequences of the selected balls, all of which we assume are equally likely.

Suppose there are $i$ red balls among the $6$ selected. If no red balls are adjacent, then we must have $0 le i le 3$, and the positions of the red balls can be chosen in $binom6-i+1i$ ways. Once the positions of the $i$ red balls are chosen, they can be selected and sequenced in $binom10i i!$ ways, and the $6-i$ blue balls can be selected and sequenced in $binom86-i (6-i)! $ ways. So the total number of sequences in which no red balls are adjacent is
$$N = sum_i=0^3 binom6-i+1i ; binom10i i! ; binom86-i (6-i)!$$
and the probability that no red balls are adjacent is
$$p = fracN binom186 6! = 0.217195$$
So the probability that at least two red balls are adjacent is $1-p = boxed0.782805.
$

share|cite|improve this answer

edited Jul 28 at 14:34

answered Jul 24 at 17:16

awkward

5,12111021

add a comment | 

up vote
1
down vote

up vote
1
down vote

It may be easier to start by finding the probability that no red balls are adjacent.

There are $binom186 6!$ possible sequences of the selected balls, all of which we assume are equally likely.

Suppose there are $i$ red balls among the $6$ selected. If no red balls are adjacent, then we must have $0 le i le 3$, and the positions of the red balls can be chosen in $binom6-i+1i$ ways. Once the positions of the $i$ red balls are chosen, they can be selected and sequenced in $binom10i i!$ ways, and the $6-i$ blue balls can be selected and sequenced in $binom86-i (6-i)! $ ways. So the total number of sequences in which no red balls are adjacent is
$$N = sum_i=0^3 binom6-i+1i ; binom10i i! ; binom86-i (6-i)!$$
and the probability that no red balls are adjacent is
$$p = fracN binom186 6! = 0.217195$$
So the probability that at least two red balls are adjacent is $1-p = boxed0.782805.
$

share|cite|improve this answer

edited Jul 28 at 14:34

answered Jul 24 at 17:16

awkward

5,12111021

It may be easier to start by finding the probability that no red balls are adjacent.

There are $binom186 6!$ possible sequences of the selected balls, all of which we assume are equally likely.

Suppose there are $i$ red balls among the $6$ selected. If no red balls are adjacent, then we must have $0 le i le 3$, and the positions of the red balls can be chosen in $binom6-i+1i$ ways. Once the positions of the $i$ red balls are chosen, they can be selected and sequenced in $binom10i i!$ ways, and the $6-i$ blue balls can be selected and sequenced in $binom86-i (6-i)! $ ways. So the total number of sequences in which no red balls are adjacent is
$$N = sum_i=0^3 binom6-i+1i ; binom10i i! ; binom86-i (6-i)!$$
and the probability that no red balls are adjacent is
$$p = fracN binom186 6! = 0.217195$$
So the probability that at least two red balls are adjacent is $1-p = boxed0.782805.
$

share|cite|improve this answer

edited Jul 28 at 14:34

answered Jul 24 at 17:16

awkward

5,12111021

share|cite|improve this answer

share|cite|improve this answer

edited Jul 28 at 14:34

edited Jul 28 at 14:34

edited Jul 28 at 14:34

answered Jul 24 at 17:16

awkward

5,12111021

answered Jul 24 at 17:16

awkward

5,12111021

answered Jul 24 at 17:16

awkward

5,12111021

5,12111021

add a comment | 

add a comment | 

up vote
0
down vote

I would attempt this problem using the method of complementary set. Essentially, I try to find the number of ways of drawing 6 balls such that 2 consecutive red balls do not appear, and subtract this probability from 1

Now, if I draw a red ball at a particular point, the next ball I draw must be a blue ball (so that no two consecutive red balls are drawn)

Using casework, we can consider the possibilities for each ball (either red or blue), which is not as difficult as it may seem given that after a red ball there must be a blue ball

Can you continue from here?

share|cite|improve this answer

answered Jul 24 at 2:20

Vikram

1716

add a comment | 

up vote
0
down vote

I would attempt this problem using the method of complementary set. Essentially, I try to find the number of ways of drawing 6 balls such that 2 consecutive red balls do not appear, and subtract this probability from 1

Now, if I draw a red ball at a particular point, the next ball I draw must be a blue ball (so that no two consecutive red balls are drawn)

Using casework, we can consider the possibilities for each ball (either red or blue), which is not as difficult as it may seem given that after a red ball there must be a blue ball

Can you continue from here?

share|cite|improve this answer

answered Jul 24 at 2:20

Vikram

1716

add a comment | 

up vote
0
down vote

up vote
0
down vote

I would attempt this problem using the method of complementary set. Essentially, I try to find the number of ways of drawing 6 balls such that 2 consecutive red balls do not appear, and subtract this probability from 1

Now, if I draw a red ball at a particular point, the next ball I draw must be a blue ball (so that no two consecutive red balls are drawn)

Using casework, we can consider the possibilities for each ball (either red or blue), which is not as difficult as it may seem given that after a red ball there must be a blue ball

Can you continue from here?

share|cite|improve this answer

answered Jul 24 at 2:20

Vikram

1716

I would attempt this problem using the method of complementary set. Essentially, I try to find the number of ways of drawing 6 balls such that 2 consecutive red balls do not appear, and subtract this probability from 1

Now, if I draw a red ball at a particular point, the next ball I draw must be a blue ball (so that no two consecutive red balls are drawn)

Using casework, we can consider the possibilities for each ball (either red or blue), which is not as difficult as it may seem given that after a red ball there must be a blue ball

Can you continue from here?

share|cite|improve this answer

answered Jul 24 at 2:20

Vikram

1716

share|cite|improve this answer

share|cite|improve this answer

answered Jul 24 at 2:20

Vikram

1716

answered Jul 24 at 2:20

Vikram

1716

answered Jul 24 at 2:20

Vikram

1716

1716

add a comment | 

add a comment | 

up vote
0
down vote

The six balls that you draw in terms of Red and Blue are

Case 1: R - 6, B - 0: The probability that such a sequence of draw $P_1 = frac10choose68choose018choose6$

Case 2: R - 5, B - 1 : $P_2 = frac10choose58choose118choose6$

Case 3: R - 4, B - 2 : $P_3 = frac10choose48choose218choose6$

Case 4: R - 3, B - 3 : $P_4 = frac10choose38choose318choose6$

Case 5: R - 2, B - 4 : $P_5 = frac10choose28choose418choose6$

Case 6: R - 1, B - 5

Case 7: R - 0, B - 6

In cases 1, 2, 3, there at some point you would have two consecutive reds drawn.

Only in cases 4 and 5 do you have times when you don't have two consecutive reds and the probability of that would be $PP_4= frac4choose36choose3$ and in case 5,$PP_5= frac5choose26choose2$and 1 minus that will be the probability that you will find atleast one time two consecutive red. In case 4, it is $1-frac15$ and in case 5, it is $1-frac1015$. IN the last two cases there are no ways to have two consecutive reds.

Thus the required probability $= P_1+P_2+P_3+P_4times(1-PP_4)+P_5times(1-PP_5).=0.78280543$

share|cite|improve this answer

edited Jul 24 at 7:29

answered Jul 24 at 6:13

Satish Ramanathan

8,84431123

add a comment | 

up vote
0
down vote

The six balls that you draw in terms of Red and Blue are

Case 1: R - 6, B - 0: The probability that such a sequence of draw $P_1 = frac10choose68choose018choose6$

Case 2: R - 5, B - 1 : $P_2 = frac10choose58choose118choose6$

Case 3: R - 4, B - 2 : $P_3 = frac10choose48choose218choose6$

Case 4: R - 3, B - 3 : $P_4 = frac10choose38choose318choose6$

Case 5: R - 2, B - 4 : $P_5 = frac10choose28choose418choose6$

Case 6: R - 1, B - 5

Case 7: R - 0, B - 6

In cases 1, 2, 3, there at some point you would have two consecutive reds drawn.

Only in cases 4 and 5 do you have times when you don't have two consecutive reds and the probability of that would be $PP_4= frac4choose36choose3$ and in case 5,$PP_5= frac5choose26choose2$and 1 minus that will be the probability that you will find atleast one time two consecutive red. In case 4, it is $1-frac15$ and in case 5, it is $1-frac1015$. IN the last two cases there are no ways to have two consecutive reds.

Thus the required probability $= P_1+P_2+P_3+P_4times(1-PP_4)+P_5times(1-PP_5).=0.78280543$

share|cite|improve this answer

edited Jul 24 at 7:29

answered Jul 24 at 6:13

Satish Ramanathan

8,84431123

add a comment | 

up vote
0
down vote

up vote
0
down vote

The six balls that you draw in terms of Red and Blue are

Case 1: R - 6, B - 0: The probability that such a sequence of draw $P_1 = frac10choose68choose018choose6$

Case 2: R - 5, B - 1 : $P_2 = frac10choose58choose118choose6$

Case 3: R - 4, B - 2 : $P_3 = frac10choose48choose218choose6$

Case 4: R - 3, B - 3 : $P_4 = frac10choose38choose318choose6$

Case 5: R - 2, B - 4 : $P_5 = frac10choose28choose418choose6$

Case 6: R - 1, B - 5

Case 7: R - 0, B - 6

In cases 1, 2, 3, there at some point you would have two consecutive reds drawn.

Only in cases 4 and 5 do you have times when you don't have two consecutive reds and the probability of that would be $PP_4= frac4choose36choose3$ and in case 5,$PP_5= frac5choose26choose2$and 1 minus that will be the probability that you will find atleast one time two consecutive red. In case 4, it is $1-frac15$ and in case 5, it is $1-frac1015$. IN the last two cases there are no ways to have two consecutive reds.

Thus the required probability $= P_1+P_2+P_3+P_4times(1-PP_4)+P_5times(1-PP_5).=0.78280543$

share|cite|improve this answer

edited Jul 24 at 7:29

answered Jul 24 at 6:13

Satish Ramanathan

8,84431123

The six balls that you draw in terms of Red and Blue are

Case 1: R - 6, B - 0: The probability that such a sequence of draw $P_1 = frac10choose68choose018choose6$

Case 2: R - 5, B - 1 : $P_2 = frac10choose58choose118choose6$

Case 3: R - 4, B - 2 : $P_3 = frac10choose48choose218choose6$

Case 4: R - 3, B - 3 : $P_4 = frac10choose38choose318choose6$

Case 5: R - 2, B - 4 : $P_5 = frac10choose28choose418choose6$

Case 6: R - 1, B - 5

Case 7: R - 0, B - 6

In cases 1, 2, 3, there at some point you would have two consecutive reds drawn.

Only in cases 4 and 5 do you have times when you don't have two consecutive reds and the probability of that would be $PP_4= frac4choose36choose3$ and in case 5,$PP_5= frac5choose26choose2$and 1 minus that will be the probability that you will find atleast one time two consecutive red. In case 4, it is $1-frac15$ and in case 5, it is $1-frac1015$. IN the last two cases there are no ways to have two consecutive reds.

Thus the required probability $= P_1+P_2+P_3+P_4times(1-PP_4)+P_5times(1-PP_5).=0.78280543$

share|cite|improve this answer

edited Jul 24 at 7:29

answered Jul 24 at 6:13

Satish Ramanathan

8,84431123

share|cite|improve this answer

share|cite|improve this answer

edited Jul 24 at 7:29

edited Jul 24 at 7:29

edited Jul 24 at 7:29

answered Jul 24 at 6:13

Satish Ramanathan

8,84431123

answered Jul 24 at 6:13

Satish Ramanathan

8,84431123

answered Jul 24 at 6:13

Satish Ramanathan

8,84431123

8,84431123

add a comment | 

add a comment | 

 

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