A coin is flipped 6 times. What is the probability that heads and tails occur an equal number of times?
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Question from my first semester Discrete Mathematics course.
A coin is flipped 6 times. What is the probability that heads and tails occur an equal number of times?
I've figured out that there are $64$ possible outcomes ($2$ outcomes each flip, $6$ flips $= 2^6 = 64$) and that in order for there to be an equal number of heads and tails exactly $3$ heads and $3$ tails must occur.
I also think order doesn't matter, so then it would be a combination / the total possible outcomes, but I'm not sure how to set up the combination or go any further.
Thanks!
probability discrete-mathematics
edited Nov 9 '17 at 0:56
Michael Hardy
1
asked Nov 9 '17 at 0:43
CheykoCheyko
63
$endgroup$
|
show 2 more comments
$begingroup$
Question from my first semester Discrete Mathematics course.
A coin is flipped 6 times. What is the probability that heads and tails occur an equal number of times?
I've figured out that there are $64$ possible outcomes ($2$ outcomes each flip, $6$ flips $= 2^6 = 64$) and that in order for there to be an equal number of heads and tails exactly $3$ heads and $3$ tails must occur.
I also think order doesn't matter, so then it would be a combination / the total possible outcomes, but I'm not sure how to set up the combination or go any further.
Thanks!
probability discrete-mathematics
edited Nov 9 '17 at 0:56
Michael Hardy
1
asked Nov 9 '17 at 0:43
CheykoCheyko
63
$endgroup$
$begingroup$
Not "at least". You require exactly 3 heads and 3 tails.
$endgroup$
– Graham Kemp
Nov 9 '17 at 0:45
$begingroup$
lol sorry. Fixed the wording.
$endgroup$
– Cheyko
Nov 9 '17 at 0:47
1
$begingroup$
it sounds like you are all the way there. you have the correct denominator. It is a combinations problem. You have idenified how many of each you need to choose. So what is the problem?
$endgroup$
– Doug M
Nov 9 '17 at 0:47
$begingroup$
I'm not sure how to set up the combination problem. I know if it was 3 heads then it would be out of 6 possibilities, take 3... 6!/3!3! and the same for 3 tails, but I don't know how to put it together... 3 heads AND 3 tails.
$endgroup$
– Cheyko
Nov 9 '17 at 0:50
1
$begingroup$
Okay, I went ahead and did the 6!/3!3! and got 20; then I took the 20 and divided it by the total possibilities from before (64) and got 20/64 or 5/16.
$endgroup$
– Cheyko
Nov 9 '17 at 0:56
|
show 2 more comments
$begingroup$
Question from my first semester Discrete Mathematics course.
A coin is flipped 6 times. What is the probability that heads and tails occur an equal number of times?
I've figured out that there are $64$ possible outcomes ($2$ outcomes each flip, $6$ flips $= 2^6 = 64$) and that in order for there to be an equal number of heads and tails exactly $3$ heads and $3$ tails must occur.
I also think order doesn't matter, so then it would be a combination / the total possible outcomes, but I'm not sure how to set up the combination or go any further.
Thanks!
probability discrete-mathematics
edited Nov 9 '17 at 0:56
Michael Hardy
1
asked Nov 9 '17 at 0:43
CheykoCheyko
63
$endgroup$
Question from my first semester Discrete Mathematics course.
A coin is flipped 6 times. What is the probability that heads and tails occur an equal number of times?
I've figured out that there are $64$ possible outcomes ($2$ outcomes each flip, $6$ flips $= 2^6 = 64$) and that in order for there to be an equal number of heads and tails exactly $3$ heads and $3$ tails must occur.
I also think order doesn't matter, so then it would be a combination / the total possible outcomes, but I'm not sure how to set up the combination or go any further.
Thanks!
probability discrete-mathematics
probability discrete-mathematics
edited Nov 9 '17 at 0:56
Michael Hardy
1
asked Nov 9 '17 at 0:43
CheykoCheyko
63
edited Nov 9 '17 at 0:56
Michael Hardy
1
asked Nov 9 '17 at 0:43
CheykoCheyko
63
edited Nov 9 '17 at 0:56
Michael Hardy
1
edited Nov 9 '17 at 0:56
Michael Hardy
1
edited Nov 9 '17 at 0:56
Michael Hardy
1
1
asked Nov 9 '17 at 0:43
CheykoCheyko
63
asked Nov 9 '17 at 0:43
CheykoCheyko
63
asked Nov 9 '17 at 0:43
CheykoCheyko
63
63
$begingroup$
Not "at least". You require exactly 3 heads and 3 tails.
$endgroup$
– Graham Kemp
Nov 9 '17 at 0:45
$begingroup$
lol sorry. Fixed the wording.
$endgroup$
– Cheyko
Nov 9 '17 at 0:47
1
$begingroup$
it sounds like you are all the way there. you have the correct denominator. It is a combinations problem. You have idenified how many of each you need to choose. So what is the problem?
$endgroup$
– Doug M
Nov 9 '17 at 0:47
$begingroup$
I'm not sure how to set up the combination problem. I know if it was 3 heads then it would be out of 6 possibilities, take 3... 6!/3!3! and the same for 3 tails, but I don't know how to put it together... 3 heads AND 3 tails.
$endgroup$
– Cheyko
Nov 9 '17 at 0:50
1
$begingroup$
Okay, I went ahead and did the 6!/3!3! and got 20; then I took the 20 and divided it by the total possibilities from before (64) and got 20/64 or 5/16.
$endgroup$
– Cheyko
Nov 9 '17 at 0:56
|
show 2 more comments
$begingroup$
Not "at least". You require exactly 3 heads and 3 tails.
$endgroup$
– Graham Kemp
Nov 9 '17 at 0:45
$begingroup$
lol sorry. Fixed the wording.
$endgroup$
– Cheyko
Nov 9 '17 at 0:47
1
$begingroup$
it sounds like you are all the way there. you have the correct denominator. It is a combinations problem. You have idenified how many of each you need to choose. So what is the problem?
$endgroup$
– Doug M
Nov 9 '17 at 0:47
$begingroup$
I'm not sure how to set up the combination problem. I know if it was 3 heads then it would be out of 6 possibilities, take 3... 6!/3!3! and the same for 3 tails, but I don't know how to put it together... 3 heads AND 3 tails.
$endgroup$
– Cheyko
Nov 9 '17 at 0:50
1
$begingroup$
Okay, I went ahead and did the 6!/3!3! and got 20; then I took the 20 and divided it by the total possibilities from before (64) and got 20/64 or 5/16.
$endgroup$
– Cheyko
Nov 9 '17 at 0:56
$begingroup$
Not "at least". You require exactly 3 heads and 3 tails.
$endgroup$
– Graham Kemp
Nov 9 '17 at 0:45
$begingroup$
Not "at least". You require exactly 3 heads and 3 tails.
$endgroup$
– Graham Kemp
Nov 9 '17 at 0:45
$begingroup$
lol sorry. Fixed the wording.
$endgroup$
– Cheyko
Nov 9 '17 at 0:47
$begingroup$
lol sorry. Fixed the wording.
$endgroup$
– Cheyko
Nov 9 '17 at 0:47
1
1
$begingroup$
it sounds like you are all the way there. you have the correct denominator. It is a combinations problem. You have idenified how many of each you need to choose. So what is the problem?
$endgroup$
– Doug M
Nov 9 '17 at 0:47
$begingroup$
it sounds like you are all the way there. you have the correct denominator. It is a combinations problem. You have idenified how many of each you need to choose. So what is the problem?
$endgroup$
– Doug M
Nov 9 '17 at 0:47
$begingroup$
I'm not sure how to set up the combination problem. I know if it was 3 heads then it would be out of 6 possibilities, take 3... 6!/3!3! and the same for 3 tails, but I don't know how to put it together... 3 heads AND 3 tails.
$endgroup$
– Cheyko
Nov 9 '17 at 0:50
$begingroup$
I'm not sure how to set up the combination problem. I know if it was 3 heads then it would be out of 6 possibilities, take 3... 6!/3!3! and the same for 3 tails, but I don't know how to put it together... 3 heads AND 3 tails.
$endgroup$
– Cheyko
Nov 9 '17 at 0:50
1
1
$begingroup$
Okay, I went ahead and did the 6!/3!3! and got 20; then I took the 20 and divided it by the total possibilities from before (64) and got 20/64 or 5/16.
$endgroup$
– Cheyko
Nov 9 '17 at 0:56
$begingroup$
Okay, I went ahead and did the 6!/3!3! and got 20; then I took the 20 and divided it by the total possibilities from before (64) and got 20/64 or 5/16.
$endgroup$
– Cheyko
Nov 9 '17 at 0:56
|
show 2 more comments
4 Answers
4
active
oldest
votes
$begingroup$
You need to count the arrangements for exactly 3 heads and 3 tails.
That is the permutations of $sf HHHTTT$.
As you state in subsequent comments, that is $6!/3!^2$.
$endgroup$
$begingroup$
Does the fact that it can be HTHTHT or any other variation of 3 heads and 3 tails not change the calculation at all?
$endgroup$
– Cheyko
Nov 9 '17 at 0:48
$begingroup$
@Cheyko counting the variations is exactly what you need to do.
$endgroup$
– Doug M
Nov 9 '17 at 0:48
$begingroup$
Yes it does: as stated, you need to count these arangements.
$endgroup$
– Graham Kemp
Nov 9 '17 at 0:51
$begingroup$
I'm sorry, I don't know what it means to 'count those arrangements'. Is that a permutation or a combination problem?
$endgroup$
– Cheyko
Nov 9 '17 at 0:53
$begingroup$
Which part of "You need to count the permutations of" is unclear?
$endgroup$
– Graham Kemp
Nov 9 '17 at 0:57
|
show 1 more comment
$begingroup$
As the other answer has said you need to count the number of permutations of $displaystyle HHHTTT$. Keep in mind that there are two groups of indistinguishable items (outcomes).
The total number of permutations of six dissimilar objects is $displaystyle 6! = 720$.
When there are two groups comprising $3$ identical objects, the number of permutations becomes: $displaystyle frac{720}{3!3!} = 20$.
If you divide this by the total number of possibilities in the event space which you've already figured out ($2^6 = 64$), you get the required probability as $displaystyle frac{20}{64} = frac{5}{16}$.
That's the combinatorics approach. If you're not limited in the way you can solve this, I would just use the concept of Bernoulli trials here. The probability you're looking for is exactly the middle term of the binomial sum of $displaystyle (frac 12 + frac 12)^6$, which is $displaystyle binom 63 (frac 12)^3(frac 12)^3 = frac{5}{16}$ as before.
answered Nov 9 '17 at 1:00
DeepakDeepak
18k11641
$endgroup$
add a comment |
$begingroup$
In general, there are $$n choose k$$ ways to get $k$ heads with $n$ coin flips. The $n$-th row in Pascal's Triangle gives those values for $k$ ranging from $0$ to $k$. So, one way to find the probability of getting $k$ heads with $n$ flips is to divide $n choose k$ by the sum of all these values ... which turns out to be $2^n$ ... and so the probability is:
$$frac{n choose k}{2^n}$$
answered Nov 9 '17 at 1:05
Bram28Bram28
64.7k44793
$endgroup$
add a comment |
$begingroup$
This is equivalent to asking what's the probability of getting exactly 3 heads in 6 tosses of a fair coin. Using the binomial distribution formula, the answer is $binom{6}{3} (1/2)^3 (1/2)^3 = binom{6}{3}/2^6$.
answered Dec 24 '18 at 10:11
littleOlittleO
30.7k649111
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You need to count the arrangements for exactly 3 heads and 3 tails.
That is the permutations of $sf HHHTTT$.
As you state in subsequent comments, that is $6!/3!^2$.
$endgroup$
$begingroup$
Does the fact that it can be HTHTHT or any other variation of 3 heads and 3 tails not change the calculation at all?
$endgroup$
– Cheyko
Nov 9 '17 at 0:48
$begingroup$
@Cheyko counting the variations is exactly what you need to do.
$endgroup$
– Doug M
Nov 9 '17 at 0:48
$begingroup$
Yes it does: as stated, you need to count these arangements.
$endgroup$
– Graham Kemp
Nov 9 '17 at 0:51
$begingroup$
I'm sorry, I don't know what it means to 'count those arrangements'. Is that a permutation or a combination problem?
$endgroup$
– Cheyko
Nov 9 '17 at 0:53
$begingroup$
Which part of "You need to count the permutations of" is unclear?
$endgroup$
– Graham Kemp
Nov 9 '17 at 0:57
|
show 1 more comment
$begingroup$
You need to count the arrangements for exactly 3 heads and 3 tails.
That is the permutations of $sf HHHTTT$.
As you state in subsequent comments, that is $6!/3!^2$.
$endgroup$
$begingroup$
Does the fact that it can be HTHTHT or any other variation of 3 heads and 3 tails not change the calculation at all?
$endgroup$
– Cheyko
Nov 9 '17 at 0:48
$begingroup$
@Cheyko counting the variations is exactly what you need to do.
$endgroup$
– Doug M
Nov 9 '17 at 0:48
$begingroup$
Yes it does: as stated, you need to count these arangements.
$endgroup$
– Graham Kemp
Nov 9 '17 at 0:51
$begingroup$
I'm sorry, I don't know what it means to 'count those arrangements'. Is that a permutation or a combination problem?
$endgroup$
– Cheyko
Nov 9 '17 at 0:53
$begingroup$
Which part of "You need to count the permutations of" is unclear?
$endgroup$
– Graham Kemp
Nov 9 '17 at 0:57
|
show 1 more comment
$begingroup$
You need to count the arrangements for exactly 3 heads and 3 tails.
That is the permutations of $sf HHHTTT$.
As you state in subsequent comments, that is $6!/3!^2$.
$endgroup$
You need to count the arrangements for exactly 3 heads and 3 tails.
That is the permutations of $sf HHHTTT$.
As you state in subsequent comments, that is $6!/3!^2$.
edited Nov 9 '17 at 0:56
community wiki
2 revs
Graham Kemp
$begingroup$
Does the fact that it can be HTHTHT or any other variation of 3 heads and 3 tails not change the calculation at all?
$endgroup$
– Cheyko
Nov 9 '17 at 0:48
$begingroup$
@Cheyko counting the variations is exactly what you need to do.
$endgroup$
– Doug M
Nov 9 '17 at 0:48
$begingroup$
Yes it does: as stated, you need to count these arangements.
$endgroup$
– Graham Kemp
Nov 9 '17 at 0:51
$begingroup$
I'm sorry, I don't know what it means to 'count those arrangements'. Is that a permutation or a combination problem?
$endgroup$
– Cheyko
Nov 9 '17 at 0:53
$begingroup$
Which part of "You need to count the permutations of" is unclear?
$endgroup$
– Graham Kemp
Nov 9 '17 at 0:57
|
show 1 more comment
$begingroup$
Does the fact that it can be HTHTHT or any other variation of 3 heads and 3 tails not change the calculation at all?
$endgroup$
– Cheyko
Nov 9 '17 at 0:48
$begingroup$
@Cheyko counting the variations is exactly what you need to do.
$endgroup$
– Doug M
Nov 9 '17 at 0:48
$begingroup$
Yes it does: as stated, you need to count these arangements.
$endgroup$
– Graham Kemp
Nov 9 '17 at 0:51
$begingroup$
I'm sorry, I don't know what it means to 'count those arrangements'. Is that a permutation or a combination problem?
$endgroup$
– Cheyko
Nov 9 '17 at 0:53
$begingroup$
Which part of "You need to count the permutations of" is unclear?
$endgroup$
– Graham Kemp
Nov 9 '17 at 0:57
$begingroup$
Does the fact that it can be HTHTHT or any other variation of 3 heads and 3 tails not change the calculation at all?
$endgroup$
– Cheyko
Nov 9 '17 at 0:48
$begingroup$
Does the fact that it can be HTHTHT or any other variation of 3 heads and 3 tails not change the calculation at all?
$endgroup$
– Cheyko
Nov 9 '17 at 0:48
$begingroup$
@Cheyko counting the variations is exactly what you need to do.
$endgroup$
– Doug M
Nov 9 '17 at 0:48
$begingroup$
@Cheyko counting the variations is exactly what you need to do.
$endgroup$
– Doug M
Nov 9 '17 at 0:48
$begingroup$
Yes it does: as stated, you need to count these arangements.
$endgroup$
– Graham Kemp
Nov 9 '17 at 0:51
$begingroup$
Yes it does: as stated, you need to count these arangements.
$endgroup$
– Graham Kemp
Nov 9 '17 at 0:51
$begingroup$
I'm sorry, I don't know what it means to 'count those arrangements'. Is that a permutation or a combination problem?
$endgroup$
– Cheyko
Nov 9 '17 at 0:53
$begingroup$
I'm sorry, I don't know what it means to 'count those arrangements'. Is that a permutation or a combination problem?
$endgroup$
– Cheyko
Nov 9 '17 at 0:53
$begingroup$
Which part of "You need to count the permutations of" is unclear?
$endgroup$
– Graham Kemp
Nov 9 '17 at 0:57
$begingroup$
Which part of "You need to count the permutations of" is unclear?
$endgroup$
– Graham Kemp
Nov 9 '17 at 0:57
|
show 1 more comment
$begingroup$
As the other answer has said you need to count the number of permutations of $displaystyle HHHTTT$. Keep in mind that there are two groups of indistinguishable items (outcomes).
The total number of permutations of six dissimilar objects is $displaystyle 6! = 720$.
When there are two groups comprising $3$ identical objects, the number of permutations becomes: $displaystyle frac{720}{3!3!} = 20$.
If you divide this by the total number of possibilities in the event space which you've already figured out ($2^6 = 64$), you get the required probability as $displaystyle frac{20}{64} = frac{5}{16}$.
That's the combinatorics approach. If you're not limited in the way you can solve this, I would just use the concept of Bernoulli trials here. The probability you're looking for is exactly the middle term of the binomial sum of $displaystyle (frac 12 + frac 12)^6$, which is $displaystyle binom 63 (frac 12)^3(frac 12)^3 = frac{5}{16}$ as before.
answered Nov 9 '17 at 1:00
DeepakDeepak
18k11641
$endgroup$
add a comment |
$begingroup$
As the other answer has said you need to count the number of permutations of $displaystyle HHHTTT$. Keep in mind that there are two groups of indistinguishable items (outcomes).
The total number of permutations of six dissimilar objects is $displaystyle 6! = 720$.
When there are two groups comprising $3$ identical objects, the number of permutations becomes: $displaystyle frac{720}{3!3!} = 20$.
If you divide this by the total number of possibilities in the event space which you've already figured out ($2^6 = 64$), you get the required probability as $displaystyle frac{20}{64} = frac{5}{16}$.
That's the combinatorics approach. If you're not limited in the way you can solve this, I would just use the concept of Bernoulli trials here. The probability you're looking for is exactly the middle term of the binomial sum of $displaystyle (frac 12 + frac 12)^6$, which is $displaystyle binom 63 (frac 12)^3(frac 12)^3 = frac{5}{16}$ as before.
answered Nov 9 '17 at 1:00
DeepakDeepak
18k11641
$endgroup$
add a comment |
$begingroup$
As the other answer has said you need to count the number of permutations of $displaystyle HHHTTT$. Keep in mind that there are two groups of indistinguishable items (outcomes).
The total number of permutations of six dissimilar objects is $displaystyle 6! = 720$.
When there are two groups comprising $3$ identical objects, the number of permutations becomes: $displaystyle frac{720}{3!3!} = 20$.
If you divide this by the total number of possibilities in the event space which you've already figured out ($2^6 = 64$), you get the required probability as $displaystyle frac{20}{64} = frac{5}{16}$.
That's the combinatorics approach. If you're not limited in the way you can solve this, I would just use the concept of Bernoulli trials here. The probability you're looking for is exactly the middle term of the binomial sum of $displaystyle (frac 12 + frac 12)^6$, which is $displaystyle binom 63 (frac 12)^3(frac 12)^3 = frac{5}{16}$ as before.
answered Nov 9 '17 at 1:00
DeepakDeepak
18k11641
$endgroup$
As the other answer has said you need to count the number of permutations of $displaystyle HHHTTT$. Keep in mind that there are two groups of indistinguishable items (outcomes).
The total number of permutations of six dissimilar objects is $displaystyle 6! = 720$.
When there are two groups comprising $3$ identical objects, the number of permutations becomes: $displaystyle frac{720}{3!3!} = 20$.
If you divide this by the total number of possibilities in the event space which you've already figured out ($2^6 = 64$), you get the required probability as $displaystyle frac{20}{64} = frac{5}{16}$.
That's the combinatorics approach. If you're not limited in the way you can solve this, I would just use the concept of Bernoulli trials here. The probability you're looking for is exactly the middle term of the binomial sum of $displaystyle (frac 12 + frac 12)^6$, which is $displaystyle binom 63 (frac 12)^3(frac 12)^3 = frac{5}{16}$ as before.
answered Nov 9 '17 at 1:00
DeepakDeepak
18k11641
answered Nov 9 '17 at 1:00
DeepakDeepak
18k11641
answered Nov 9 '17 at 1:00
DeepakDeepak
18k11641
answered Nov 9 '17 at 1:00
DeepakDeepak
18k11641
18k11641
add a comment |
add a comment |
$begingroup$
In general, there are $$n choose k$$ ways to get $k$ heads with $n$ coin flips. The $n$-th row in Pascal's Triangle gives those values for $k$ ranging from $0$ to $k$. So, one way to find the probability of getting $k$ heads with $n$ flips is to divide $n choose k$ by the sum of all these values ... which turns out to be $2^n$ ... and so the probability is:
$$frac{n choose k}{2^n}$$
answered Nov 9 '17 at 1:05
Bram28Bram28
64.7k44793
$endgroup$
add a comment |
$begingroup$
In general, there are $$n choose k$$ ways to get $k$ heads with $n$ coin flips. The $n$-th row in Pascal's Triangle gives those values for $k$ ranging from $0$ to $k$. So, one way to find the probability of getting $k$ heads with $n$ flips is to divide $n choose k$ by the sum of all these values ... which turns out to be $2^n$ ... and so the probability is:
$$frac{n choose k}{2^n}$$
answered Nov 9 '17 at 1:05
Bram28Bram28
64.7k44793
$endgroup$
add a comment |
$begingroup$
In general, there are $$n choose k$$ ways to get $k$ heads with $n$ coin flips. The $n$-th row in Pascal's Triangle gives those values for $k$ ranging from $0$ to $k$. So, one way to find the probability of getting $k$ heads with $n$ flips is to divide $n choose k$ by the sum of all these values ... which turns out to be $2^n$ ... and so the probability is:
$$frac{n choose k}{2^n}$$
answered Nov 9 '17 at 1:05
Bram28Bram28
64.7k44793
$endgroup$
In general, there are $$n choose k$$ ways to get $k$ heads with $n$ coin flips. The $n$-th row in Pascal's Triangle gives those values for $k$ ranging from $0$ to $k$. So, one way to find the probability of getting $k$ heads with $n$ flips is to divide $n choose k$ by the sum of all these values ... which turns out to be $2^n$ ... and so the probability is:
$$frac{n choose k}{2^n}$$
answered Nov 9 '17 at 1:05
Bram28Bram28
64.7k44793
answered Nov 9 '17 at 1:05
Bram28Bram28
64.7k44793
answered Nov 9 '17 at 1:05
Bram28Bram28
64.7k44793
answered Nov 9 '17 at 1:05
Bram28Bram28
64.7k44793
64.7k44793
add a comment |
add a comment |
$begingroup$
This is equivalent to asking what's the probability of getting exactly 3 heads in 6 tosses of a fair coin. Using the binomial distribution formula, the answer is $binom{6}{3} (1/2)^3 (1/2)^3 = binom{6}{3}/2^6$.
answered Dec 24 '18 at 10:11
littleOlittleO
30.7k649111
$endgroup$
add a comment |
$begingroup$
This is equivalent to asking what's the probability of getting exactly 3 heads in 6 tosses of a fair coin. Using the binomial distribution formula, the answer is $binom{6}{3} (1/2)^3 (1/2)^3 = binom{6}{3}/2^6$.
answered Dec 24 '18 at 10:11
littleOlittleO
30.7k649111
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add a comment |
$begingroup$
This is equivalent to asking what's the probability of getting exactly 3 heads in 6 tosses of a fair coin. Using the binomial distribution formula, the answer is $binom{6}{3} (1/2)^3 (1/2)^3 = binom{6}{3}/2^6$.
answered Dec 24 '18 at 10:11
littleOlittleO
30.7k649111
$endgroup$
This is equivalent to asking what's the probability of getting exactly 3 heads in 6 tosses of a fair coin. Using the binomial distribution formula, the answer is $binom{6}{3} (1/2)^3 (1/2)^3 = binom{6}{3}/2^6$.
answered Dec 24 '18 at 10:11
littleOlittleO
30.7k649111
answered Dec 24 '18 at 10:11
littleOlittleO
30.7k649111
answered Dec 24 '18 at 10:11
littleOlittleO
30.7k649111
answered Dec 24 '18 at 10:11
littleOlittleO
30.7k649111
30.7k649111
add a comment |
add a comment |
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Not "at least". You require exactly 3 heads and 3 tails.
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– Graham Kemp
Nov 9 '17 at 0:45
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lol sorry. Fixed the wording.
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– Cheyko
Nov 9 '17 at 0:47
1
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it sounds like you are all the way there. you have the correct denominator. It is a combinations problem. You have idenified how many of each you need to choose. So what is the problem?
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– Doug M
Nov 9 '17 at 0:47
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I'm not sure how to set up the combination problem. I know if it was 3 heads then it would be out of 6 possibilities, take 3... 6!/3!3! and the same for 3 tails, but I don't know how to put it together... 3 heads AND 3 tails.
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– Cheyko
Nov 9 '17 at 0:50
1
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Okay, I went ahead and did the 6!/3!3! and got 20; then I took the 20 and divided it by the total possibilities from before (64) and got 20/64 or 5/16.
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– Cheyko
Nov 9 '17 at 0:56