A coin is flipped 6 times. What is the probability that heads and tails occur an equal number of times?












1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














1












1








1





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 9 '17 at 0:56









Michael Hardy

1




1










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


















  • $begingroup$
    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










4 Answers
4






active

oldest

votes


















0












$begingroup$

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 $6!/3!^2$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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



















0












$begingroup$

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.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    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:



    $$frac{n choose k}{2^n}$$






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      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 $binom{6}{3} (1/2)^3 (1/2)^3 = binom{6}{3}/2^6$.






      share|cite|improve this answer









      $endgroup$













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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        0












        $begingroup$

        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 $6!/3!^2$.






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          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
















        0












        $begingroup$

        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 $6!/3!^2$.






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          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














        0












        0








        0





        $begingroup$

        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 $6!/3!^2$.






        share|cite|improve this answer











        $endgroup$



        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 $6!/3!^2$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        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


















        • $begingroup$
          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











        0












        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$


















          0












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$
















            0












            0








            0





            $begingroup$

            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.






            share|cite|improve this answer









            $endgroup$



            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.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 9 '17 at 1:00









            DeepakDeepak

            17k11436




            17k11436























                0












                $begingroup$

                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:



                $$frac{n choose k}{2^n}$$






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  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:



                  $$frac{n choose k}{2^n}$$






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    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:



                    $$frac{n choose k}{2^n}$$






                    share|cite|improve this answer









                    $endgroup$



                    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:



                    $$frac{n choose k}{2^n}$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 9 '17 at 1:05









                    Bram28Bram28

                    62k44793




                    62k44793























                        0












                        $begingroup$

                        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 $binom{6}{3} (1/2)^3 (1/2)^3 = binom{6}{3}/2^6$.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          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 $binom{6}{3} (1/2)^3 (1/2)^3 = binom{6}{3}/2^6$.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            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 $binom{6}{3} (1/2)^3 (1/2)^3 = binom{6}{3}/2^6$.






                            share|cite|improve this answer









                            $endgroup$



                            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 $binom{6}{3} (1/2)^3 (1/2)^3 = binom{6}{3}/2^6$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 24 '18 at 10:11









                            littleOlittleO

                            29.8k646109




                            29.8k646109






























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