What is the probability that each element in this string is non-zero?












0












$begingroup$


Question: You are given two bitstrings a1, a2, .. a77 and b1, b2, .. b77 of length 77 In both bitstrings, each bit is 0 with probability 3/4, and 1 with probability 1/4 (independent of all other bits).



Consider the string: a1-b1, a2-b2, ..., a77-b77.



What is the probability that each element in this string is non-zero?



Answer: 1.586381421*10^-33



Attempt:



Non-zero probability is 1/4 for both bitstrings.
If both bitstrings are subtracted by one another for length 77, then it shouldn't it be (1/4)*(1/4) = 1/16 be the probability of the new string being non-zero



Total number of possible bitstrings = 2^77



Probability: 1/16/2^77 = 4.13*10^-25






probability probability-theory discrete-mathematics







share|cite|improve this question











$endgroup$












  • $begingroup$
    $a_1-b_1neq 0$ just means that $a_1neq b_1$. It certainly does not mean that both of the elements are $1$. Also, while it is true that there are $2^{77}$ possible strings, they are not equi-probable so you can't use that as a denominator in the way you have.
    $endgroup$
    – lulu
    Nov 26 '18 at 21:21












  • $begingroup$
    Here's a MathJax tutorial :)
    $endgroup$
    – Shaun
    Nov 26 '18 at 21:21










  • $begingroup$
    As a suggestion: start with a shorter string. What is the probability if both strings have length $1$? Length $2$? Length $3$?
    $endgroup$
    – lulu
    Nov 26 '18 at 21:23










  • $begingroup$
    @lulu Isn't the question basically asking to find the probability of each element being non-zero so meaning it has to be 1? Bitstrings have either 0 or 1, so according to this, non-zero bitstrings = bitstrings with only 1s?
    $endgroup$
    – Toby
    Nov 26 '18 at 21:44






  • 1




    $begingroup$
    No. The difference is not itself a bit string. if $a_1=0,b_1=1$ then $a_1-b_1=-1$ which is non-zero.
    $endgroup$
    – lulu
    Nov 26 '18 at 22:05
















0












$begingroup$


Question: You are given two bitstrings a1, a2, .. a77 and b1, b2, .. b77 of length 77 In both bitstrings, each bit is 0 with probability 3/4, and 1 with probability 1/4 (independent of all other bits).



Consider the string: a1-b1, a2-b2, ..., a77-b77.



What is the probability that each element in this string is non-zero?



Answer: 1.586381421*10^-33



Attempt:



Non-zero probability is 1/4 for both bitstrings.
If both bitstrings are subtracted by one another for length 77, then it shouldn't it be (1/4)*(1/4) = 1/16 be the probability of the new string being non-zero



Total number of possible bitstrings = 2^77



Probability: 1/16/2^77 = 4.13*10^-25






probability probability-theory discrete-mathematics







share|cite|improve this question











$endgroup$












  • $begingroup$
    $a_1-b_1neq 0$ just means that $a_1neq b_1$. It certainly does not mean that both of the elements are $1$. Also, while it is true that there are $2^{77}$ possible strings, they are not equi-probable so you can't use that as a denominator in the way you have.
    $endgroup$
    – lulu
    Nov 26 '18 at 21:21












  • $begingroup$
    Here's a MathJax tutorial :)
    $endgroup$
    – Shaun
    Nov 26 '18 at 21:21










  • $begingroup$
    As a suggestion: start with a shorter string. What is the probability if both strings have length $1$? Length $2$? Length $3$?
    $endgroup$
    – lulu
    Nov 26 '18 at 21:23










  • $begingroup$
    @lulu Isn't the question basically asking to find the probability of each element being non-zero so meaning it has to be 1? Bitstrings have either 0 or 1, so according to this, non-zero bitstrings = bitstrings with only 1s?
    $endgroup$
    – Toby
    Nov 26 '18 at 21:44






  • 1




    $begingroup$
    No. The difference is not itself a bit string. if $a_1=0,b_1=1$ then $a_1-b_1=-1$ which is non-zero.
    $endgroup$
    – lulu
    Nov 26 '18 at 22:05














0












0








0





$begingroup$


Question: You are given two bitstrings a1, a2, .. a77 and b1, b2, .. b77 of length 77 In both bitstrings, each bit is 0 with probability 3/4, and 1 with probability 1/4 (independent of all other bits).



Consider the string: a1-b1, a2-b2, ..., a77-b77.



What is the probability that each element in this string is non-zero?



Answer: 1.586381421*10^-33



Attempt:



Non-zero probability is 1/4 for both bitstrings.
If both bitstrings are subtracted by one another for length 77, then it shouldn't it be (1/4)*(1/4) = 1/16 be the probability of the new string being non-zero



Total number of possible bitstrings = 2^77



Probability: 1/16/2^77 = 4.13*10^-25






probability probability-theory discrete-mathematics







share|cite|improve this question











$endgroup$




Question: You are given two bitstrings a1, a2, .. a77 and b1, b2, .. b77 of length 77 In both bitstrings, each bit is 0 with probability 3/4, and 1 with probability 1/4 (independent of all other bits).



Consider the string: a1-b1, a2-b2, ..., a77-b77.



What is the probability that each element in this string is non-zero?



Answer: 1.586381421*10^-33



Attempt:



Non-zero probability is 1/4 for both bitstrings.
If both bitstrings are subtracted by one another for length 77, then it shouldn't it be (1/4)*(1/4) = 1/16 be the probability of the new string being non-zero



Total number of possible bitstrings = 2^77



Probability: 1/16/2^77 = 4.13*10^-25






probability probability-theory discrete-mathematics




probability probability-theory discrete-mathematics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 26 '18 at 21:21









Shaun

8,939113681




8,939113681










asked Nov 26 '18 at 21:18









TobyToby

1577




1577












  • $begingroup$
    $a_1-b_1neq 0$ just means that $a_1neq b_1$. It certainly does not mean that both of the elements are $1$. Also, while it is true that there are $2^{77}$ possible strings, they are not equi-probable so you can't use that as a denominator in the way you have.
    $endgroup$
    – lulu
    Nov 26 '18 at 21:21












  • $begingroup$
    Here's a MathJax tutorial :)
    $endgroup$
    – Shaun
    Nov 26 '18 at 21:21










  • $begingroup$
    As a suggestion: start with a shorter string. What is the probability if both strings have length $1$? Length $2$? Length $3$?
    $endgroup$
    – lulu
    Nov 26 '18 at 21:23










  • $begingroup$
    @lulu Isn't the question basically asking to find the probability of each element being non-zero so meaning it has to be 1? Bitstrings have either 0 or 1, so according to this, non-zero bitstrings = bitstrings with only 1s?
    $endgroup$
    – Toby
    Nov 26 '18 at 21:44






  • 1




    $begingroup$
    No. The difference is not itself a bit string. if $a_1=0,b_1=1$ then $a_1-b_1=-1$ which is non-zero.
    $endgroup$
    – lulu
    Nov 26 '18 at 22:05


















  • $begingroup$
    $a_1-b_1neq 0$ just means that $a_1neq b_1$. It certainly does not mean that both of the elements are $1$. Also, while it is true that there are $2^{77}$ possible strings, they are not equi-probable so you can't use that as a denominator in the way you have.
    $endgroup$
    – lulu
    Nov 26 '18 at 21:21












  • $begingroup$
    Here's a MathJax tutorial :)
    $endgroup$
    – Shaun
    Nov 26 '18 at 21:21










  • $begingroup$
    As a suggestion: start with a shorter string. What is the probability if both strings have length $1$? Length $2$? Length $3$?
    $endgroup$
    – lulu
    Nov 26 '18 at 21:23










  • $begingroup$
    @lulu Isn't the question basically asking to find the probability of each element being non-zero so meaning it has to be 1? Bitstrings have either 0 or 1, so according to this, non-zero bitstrings = bitstrings with only 1s?
    $endgroup$
    – Toby
    Nov 26 '18 at 21:44






  • 1




    $begingroup$
    No. The difference is not itself a bit string. if $a_1=0,b_1=1$ then $a_1-b_1=-1$ which is non-zero.
    $endgroup$
    – lulu
    Nov 26 '18 at 22:05
















$begingroup$
$a_1-b_1neq 0$ just means that $a_1neq b_1$. It certainly does not mean that both of the elements are $1$. Also, while it is true that there are $2^{77}$ possible strings, they are not equi-probable so you can't use that as a denominator in the way you have.
$endgroup$
– lulu
Nov 26 '18 at 21:21






$begingroup$
$a_1-b_1neq 0$ just means that $a_1neq b_1$. It certainly does not mean that both of the elements are $1$. Also, while it is true that there are $2^{77}$ possible strings, they are not equi-probable so you can't use that as a denominator in the way you have.
$endgroup$
– lulu
Nov 26 '18 at 21:21














$begingroup$
Here's a MathJax tutorial :)
$endgroup$
– Shaun
Nov 26 '18 at 21:21




$begingroup$
Here's a MathJax tutorial :)
$endgroup$
– Shaun
Nov 26 '18 at 21:21












$begingroup$
As a suggestion: start with a shorter string. What is the probability if both strings have length $1$? Length $2$? Length $3$?
$endgroup$
– lulu
Nov 26 '18 at 21:23




$begingroup$
As a suggestion: start with a shorter string. What is the probability if both strings have length $1$? Length $2$? Length $3$?
$endgroup$
– lulu
Nov 26 '18 at 21:23












$begingroup$
@lulu Isn't the question basically asking to find the probability of each element being non-zero so meaning it has to be 1? Bitstrings have either 0 or 1, so according to this, non-zero bitstrings = bitstrings with only 1s?
$endgroup$
– Toby
Nov 26 '18 at 21:44




$begingroup$
@lulu Isn't the question basically asking to find the probability of each element being non-zero so meaning it has to be 1? Bitstrings have either 0 or 1, so according to this, non-zero bitstrings = bitstrings with only 1s?
$endgroup$
– Toby
Nov 26 '18 at 21:44




1




1




$begingroup$
No. The difference is not itself a bit string. if $a_1=0,b_1=1$ then $a_1-b_1=-1$ which is non-zero.
$endgroup$
– lulu
Nov 26 '18 at 22:05




$begingroup$
No. The difference is not itself a bit string. if $a_1=0,b_1=1$ then $a_1-b_1=-1$ which is non-zero.
$endgroup$
– lulu
Nov 26 '18 at 22:05










3 Answers
3






active

oldest

votes


















1












$begingroup$

Let $c_i=a_i-b_i$. You need to find $$P[c_1 neq 0, c_2 neq 0, ldots, c_{77} neq 0] = p^{77}$$



where $$p=P[c_i neq 0] = P[a_i = 0, b_i = 1] + P[a_i = 1, b_i = 0] = 2timesfrac{3}{16}=frac{6}{16}=frac{3}{8}$$



So, the final answer is $$left(frac{3}{8}right)^{77}$$ which gives the answer you provided.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Why is p to the power 77? Also, why does ai and bi have to be pairs of (0,1) and (1,0)? Is it to satisfy the ai - bi != 0 equation?
    $endgroup$
    – Toby
    Nov 26 '18 at 21:54






  • 1




    $begingroup$
    It's $p^{77}$ because the individual elements ${c_i}$ are independent (because ${a_i} text{ and }{b_i}$ are independent). Yes, $c_ineq 0$ only if $a_ineq b_i$, so, you need to consider the cases where this holds. Note that $c_ineq 0$ means that $c_i = 1$ or $c_i = -1$.
    $endgroup$
    – BlackMath
    Nov 26 '18 at 21:58





















2












$begingroup$

$a_i-b_ineq0implies a_ineq b_i$



The possible pairs $(a_i, b_i)$ are $(0, 1)$ and $(1,0)$, with each pair having a probability $frac{3}{4}cdotfrac{1}{4}=frac{3}{16}$. Therefore, the total probability is $(frac{3}{16}+frac{3}{16})^{77}=(frac{6}{16})^{77}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    why does the probability have to be to the power 77?
    $endgroup$
    – Toby
    Nov 26 '18 at 21:56






  • 1




    $begingroup$
    Because you have $77$ pairs of corresponding bits $(a_i, b_i)$, and the probability that the difference $a_i-b_ineq0$ is $6/16$ for all pairs.
    $endgroup$
    – Shubham Johri
    Nov 26 '18 at 22:02





















0












$begingroup$

The answer is $(frac{6}{16})^{77}$.
One digit i being 0 is equivalent to $lnot(a_i=b_i=0lor a_i=b_i=1)$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    can you please elaborate on how you got the answer. I don't follow your logic
    $endgroup$
    – Toby
    Nov 26 '18 at 21:39











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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Let $c_i=a_i-b_i$. You need to find $$P[c_1 neq 0, c_2 neq 0, ldots, c_{77} neq 0] = p^{77}$$



where $$p=P[c_i neq 0] = P[a_i = 0, b_i = 1] + P[a_i = 1, b_i = 0] = 2timesfrac{3}{16}=frac{6}{16}=frac{3}{8}$$



So, the final answer is $$left(frac{3}{8}right)^{77}$$ which gives the answer you provided.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Why is p to the power 77? Also, why does ai and bi have to be pairs of (0,1) and (1,0)? Is it to satisfy the ai - bi != 0 equation?
    $endgroup$
    – Toby
    Nov 26 '18 at 21:54






  • 1




    $begingroup$
    It's $p^{77}$ because the individual elements ${c_i}$ are independent (because ${a_i} text{ and }{b_i}$ are independent). Yes, $c_ineq 0$ only if $a_ineq b_i$, so, you need to consider the cases where this holds. Note that $c_ineq 0$ means that $c_i = 1$ or $c_i = -1$.
    $endgroup$
    – BlackMath
    Nov 26 '18 at 21:58


















1












$begingroup$

Let $c_i=a_i-b_i$. You need to find $$P[c_1 neq 0, c_2 neq 0, ldots, c_{77} neq 0] = p^{77}$$



where $$p=P[c_i neq 0] = P[a_i = 0, b_i = 1] + P[a_i = 1, b_i = 0] = 2timesfrac{3}{16}=frac{6}{16}=frac{3}{8}$$



So, the final answer is $$left(frac{3}{8}right)^{77}$$ which gives the answer you provided.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Why is p to the power 77? Also, why does ai and bi have to be pairs of (0,1) and (1,0)? Is it to satisfy the ai - bi != 0 equation?
    $endgroup$
    – Toby
    Nov 26 '18 at 21:54






  • 1




    $begingroup$
    It's $p^{77}$ because the individual elements ${c_i}$ are independent (because ${a_i} text{ and }{b_i}$ are independent). Yes, $c_ineq 0$ only if $a_ineq b_i$, so, you need to consider the cases where this holds. Note that $c_ineq 0$ means that $c_i = 1$ or $c_i = -1$.
    $endgroup$
    – BlackMath
    Nov 26 '18 at 21:58
















1












1








1





$begingroup$

Let $c_i=a_i-b_i$. You need to find $$P[c_1 neq 0, c_2 neq 0, ldots, c_{77} neq 0] = p^{77}$$



where $$p=P[c_i neq 0] = P[a_i = 0, b_i = 1] + P[a_i = 1, b_i = 0] = 2timesfrac{3}{16}=frac{6}{16}=frac{3}{8}$$



So, the final answer is $$left(frac{3}{8}right)^{77}$$ which gives the answer you provided.






share|cite|improve this answer









$endgroup$



Let $c_i=a_i-b_i$. You need to find $$P[c_1 neq 0, c_2 neq 0, ldots, c_{77} neq 0] = p^{77}$$



where $$p=P[c_i neq 0] = P[a_i = 0, b_i = 1] + P[a_i = 1, b_i = 0] = 2timesfrac{3}{16}=frac{6}{16}=frac{3}{8}$$



So, the final answer is $$left(frac{3}{8}right)^{77}$$ which gives the answer you provided.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 26 '18 at 21:48









BlackMathBlackMath

30518




30518












  • $begingroup$
    Why is p to the power 77? Also, why does ai and bi have to be pairs of (0,1) and (1,0)? Is it to satisfy the ai - bi != 0 equation?
    $endgroup$
    – Toby
    Nov 26 '18 at 21:54






  • 1




    $begingroup$
    It's $p^{77}$ because the individual elements ${c_i}$ are independent (because ${a_i} text{ and }{b_i}$ are independent). Yes, $c_ineq 0$ only if $a_ineq b_i$, so, you need to consider the cases where this holds. Note that $c_ineq 0$ means that $c_i = 1$ or $c_i = -1$.
    $endgroup$
    – BlackMath
    Nov 26 '18 at 21:58




















  • $begingroup$
    Why is p to the power 77? Also, why does ai and bi have to be pairs of (0,1) and (1,0)? Is it to satisfy the ai - bi != 0 equation?
    $endgroup$
    – Toby
    Nov 26 '18 at 21:54






  • 1




    $begingroup$
    It's $p^{77}$ because the individual elements ${c_i}$ are independent (because ${a_i} text{ and }{b_i}$ are independent). Yes, $c_ineq 0$ only if $a_ineq b_i$, so, you need to consider the cases where this holds. Note that $c_ineq 0$ means that $c_i = 1$ or $c_i = -1$.
    $endgroup$
    – BlackMath
    Nov 26 '18 at 21:58


















$begingroup$
Why is p to the power 77? Also, why does ai and bi have to be pairs of (0,1) and (1,0)? Is it to satisfy the ai - bi != 0 equation?
$endgroup$
– Toby
Nov 26 '18 at 21:54




$begingroup$
Why is p to the power 77? Also, why does ai and bi have to be pairs of (0,1) and (1,0)? Is it to satisfy the ai - bi != 0 equation?
$endgroup$
– Toby
Nov 26 '18 at 21:54




1




1




$begingroup$
It's $p^{77}$ because the individual elements ${c_i}$ are independent (because ${a_i} text{ and }{b_i}$ are independent). Yes, $c_ineq 0$ only if $a_ineq b_i$, so, you need to consider the cases where this holds. Note that $c_ineq 0$ means that $c_i = 1$ or $c_i = -1$.
$endgroup$
– BlackMath
Nov 26 '18 at 21:58






$begingroup$
It's $p^{77}$ because the individual elements ${c_i}$ are independent (because ${a_i} text{ and }{b_i}$ are independent). Yes, $c_ineq 0$ only if $a_ineq b_i$, so, you need to consider the cases where this holds. Note that $c_ineq 0$ means that $c_i = 1$ or $c_i = -1$.
$endgroup$
– BlackMath
Nov 26 '18 at 21:58













2












$begingroup$

$a_i-b_ineq0implies a_ineq b_i$



The possible pairs $(a_i, b_i)$ are $(0, 1)$ and $(1,0)$, with each pair having a probability $frac{3}{4}cdotfrac{1}{4}=frac{3}{16}$. Therefore, the total probability is $(frac{3}{16}+frac{3}{16})^{77}=(frac{6}{16})^{77}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    why does the probability have to be to the power 77?
    $endgroup$
    – Toby
    Nov 26 '18 at 21:56






  • 1




    $begingroup$
    Because you have $77$ pairs of corresponding bits $(a_i, b_i)$, and the probability that the difference $a_i-b_ineq0$ is $6/16$ for all pairs.
    $endgroup$
    – Shubham Johri
    Nov 26 '18 at 22:02


















2












$begingroup$

$a_i-b_ineq0implies a_ineq b_i$



The possible pairs $(a_i, b_i)$ are $(0, 1)$ and $(1,0)$, with each pair having a probability $frac{3}{4}cdotfrac{1}{4}=frac{3}{16}$. Therefore, the total probability is $(frac{3}{16}+frac{3}{16})^{77}=(frac{6}{16})^{77}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    why does the probability have to be to the power 77?
    $endgroup$
    – Toby
    Nov 26 '18 at 21:56






  • 1




    $begingroup$
    Because you have $77$ pairs of corresponding bits $(a_i, b_i)$, and the probability that the difference $a_i-b_ineq0$ is $6/16$ for all pairs.
    $endgroup$
    – Shubham Johri
    Nov 26 '18 at 22:02
















2












2








2





$begingroup$

$a_i-b_ineq0implies a_ineq b_i$



The possible pairs $(a_i, b_i)$ are $(0, 1)$ and $(1,0)$, with each pair having a probability $frac{3}{4}cdotfrac{1}{4}=frac{3}{16}$. Therefore, the total probability is $(frac{3}{16}+frac{3}{16})^{77}=(frac{6}{16})^{77}$.






share|cite|improve this answer









$endgroup$



$a_i-b_ineq0implies a_ineq b_i$



The possible pairs $(a_i, b_i)$ are $(0, 1)$ and $(1,0)$, with each pair having a probability $frac{3}{4}cdotfrac{1}{4}=frac{3}{16}$. Therefore, the total probability is $(frac{3}{16}+frac{3}{16})^{77}=(frac{6}{16})^{77}$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 26 '18 at 21:47









Shubham JohriShubham Johri

4,992717




4,992717












  • $begingroup$
    why does the probability have to be to the power 77?
    $endgroup$
    – Toby
    Nov 26 '18 at 21:56






  • 1




    $begingroup$
    Because you have $77$ pairs of corresponding bits $(a_i, b_i)$, and the probability that the difference $a_i-b_ineq0$ is $6/16$ for all pairs.
    $endgroup$
    – Shubham Johri
    Nov 26 '18 at 22:02




















  • $begingroup$
    why does the probability have to be to the power 77?
    $endgroup$
    – Toby
    Nov 26 '18 at 21:56






  • 1




    $begingroup$
    Because you have $77$ pairs of corresponding bits $(a_i, b_i)$, and the probability that the difference $a_i-b_ineq0$ is $6/16$ for all pairs.
    $endgroup$
    – Shubham Johri
    Nov 26 '18 at 22:02


















$begingroup$
why does the probability have to be to the power 77?
$endgroup$
– Toby
Nov 26 '18 at 21:56




$begingroup$
why does the probability have to be to the power 77?
$endgroup$
– Toby
Nov 26 '18 at 21:56




1




1




$begingroup$
Because you have $77$ pairs of corresponding bits $(a_i, b_i)$, and the probability that the difference $a_i-b_ineq0$ is $6/16$ for all pairs.
$endgroup$
– Shubham Johri
Nov 26 '18 at 22:02






$begingroup$
Because you have $77$ pairs of corresponding bits $(a_i, b_i)$, and the probability that the difference $a_i-b_ineq0$ is $6/16$ for all pairs.
$endgroup$
– Shubham Johri
Nov 26 '18 at 22:02













0












$begingroup$

The answer is $(frac{6}{16})^{77}$.
One digit i being 0 is equivalent to $lnot(a_i=b_i=0lor a_i=b_i=1)$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    can you please elaborate on how you got the answer. I don't follow your logic
    $endgroup$
    – Toby
    Nov 26 '18 at 21:39
















0












$begingroup$

The answer is $(frac{6}{16})^{77}$.
One digit i being 0 is equivalent to $lnot(a_i=b_i=0lor a_i=b_i=1)$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    can you please elaborate on how you got the answer. I don't follow your logic
    $endgroup$
    – Toby
    Nov 26 '18 at 21:39














0












0








0





$begingroup$

The answer is $(frac{6}{16})^{77}$.
One digit i being 0 is equivalent to $lnot(a_i=b_i=0lor a_i=b_i=1)$






share|cite|improve this answer









$endgroup$



The answer is $(frac{6}{16})^{77}$.
One digit i being 0 is equivalent to $lnot(a_i=b_i=0lor a_i=b_i=1)$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 26 '18 at 21:28









Martin ErhardtMartin Erhardt

21019




21019












  • $begingroup$
    can you please elaborate on how you got the answer. I don't follow your logic
    $endgroup$
    – Toby
    Nov 26 '18 at 21:39


















  • $begingroup$
    can you please elaborate on how you got the answer. I don't follow your logic
    $endgroup$
    – Toby
    Nov 26 '18 at 21:39
















$begingroup$
can you please elaborate on how you got the answer. I don't follow your logic
$endgroup$
– Toby
Nov 26 '18 at 21:39




$begingroup$
can you please elaborate on how you got the answer. I don't follow your logic
$endgroup$
– Toby
Nov 26 '18 at 21:39


















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