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












0














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
























  • $a_1-b_1neq 0$ 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.
    – lulu
    Nov 26 at 21:21












  • Here's a MathJax tutorial :)
    – Shaun
    Nov 26 at 21:21










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










  • @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?
    – Toby
    Nov 26 at 21:44






  • 1




    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.
    – lulu
    Nov 26 at 22:05
















0














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
























  • $a_1-b_1neq 0$ 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.
    – lulu
    Nov 26 at 21:21












  • Here's a MathJax tutorial :)
    – Shaun
    Nov 26 at 21:21










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










  • @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?
    – Toby
    Nov 26 at 21:44






  • 1




    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.
    – lulu
    Nov 26 at 22:05














0












0








0







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















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 at 21:21









Shaun

8,627113680




8,627113680










asked Nov 26 at 21:18









Toby

1577




1577












  • $a_1-b_1neq 0$ 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.
    – lulu
    Nov 26 at 21:21












  • Here's a MathJax tutorial :)
    – Shaun
    Nov 26 at 21:21










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










  • @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?
    – Toby
    Nov 26 at 21:44






  • 1




    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.
    – lulu
    Nov 26 at 22:05


















  • $a_1-b_1neq 0$ 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.
    – lulu
    Nov 26 at 21:21












  • Here's a MathJax tutorial :)
    – Shaun
    Nov 26 at 21:21










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










  • @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?
    – Toby
    Nov 26 at 21:44






  • 1




    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.
    – lulu
    Nov 26 at 22:05
















$a_1-b_1neq 0$ 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.
– lulu
Nov 26 at 21:21






$a_1-b_1neq 0$ 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.
– lulu
Nov 26 at 21:21














Here's a MathJax tutorial :)
– Shaun
Nov 26 at 21:21




Here's a MathJax tutorial :)
– Shaun
Nov 26 at 21:21












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




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












@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?
– Toby
Nov 26 at 21:44




@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?
– Toby
Nov 26 at 21:44




1




1




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.
– lulu
Nov 26 at 22:05




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.
– lulu
Nov 26 at 22:05










3 Answers
3






active

oldest

votes


















1














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





















  • 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?
    – Toby
    Nov 26 at 21:54






  • 1




    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 $c_i = -1$.
    – BlackMath
    Nov 26 at 21:58





















2














$a_i-b_ineq0implies a_ineq b_i$



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





















  • why does the probability have to be to the power 77?
    – Toby
    Nov 26 at 21:56






  • 1




    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.
    – Shubham Johri
    Nov 26 at 22:02





















0














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





















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











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3014942%2fwhat-is-the-probability-that-each-element-in-this-string-is-non-zero%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














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





















  • 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?
    – Toby
    Nov 26 at 21:54






  • 1




    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 $c_i = -1$.
    – BlackMath
    Nov 26 at 21:58


















1














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





















  • 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?
    – Toby
    Nov 26 at 21:54






  • 1




    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 $c_i = -1$.
    – BlackMath
    Nov 26 at 21:58
















1












1








1






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












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 at 21:48









BlackMath

18018




18018












  • 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?
    – Toby
    Nov 26 at 21:54






  • 1




    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 $c_i = -1$.
    – BlackMath
    Nov 26 at 21:58




















  • 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?
    – Toby
    Nov 26 at 21:54






  • 1




    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 $c_i = -1$.
    – BlackMath
    Nov 26 at 21:58


















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?
– Toby
Nov 26 at 21:54




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?
– Toby
Nov 26 at 21:54




1




1




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 $c_i = -1$.
– BlackMath
Nov 26 at 21:58






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 $c_i = -1$.
– BlackMath
Nov 26 at 21:58













2














$a_i-b_ineq0implies a_ineq b_i$



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





















  • why does the probability have to be to the power 77?
    – Toby
    Nov 26 at 21:56






  • 1




    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.
    – Shubham Johri
    Nov 26 at 22:02


















2














$a_i-b_ineq0implies a_ineq b_i$



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





















  • why does the probability have to be to the power 77?
    – Toby
    Nov 26 at 21:56






  • 1




    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.
    – Shubham Johri
    Nov 26 at 22:02
















2












2








2






$a_i-b_ineq0implies a_ineq b_i$



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












$a_i-b_ineq0implies a_ineq b_i$



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 at 21:47









Shubham Johri

3,826716




3,826716












  • why does the probability have to be to the power 77?
    – Toby
    Nov 26 at 21:56






  • 1




    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.
    – Shubham Johri
    Nov 26 at 22:02




















  • why does the probability have to be to the power 77?
    – Toby
    Nov 26 at 21:56






  • 1




    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.
    – Shubham Johri
    Nov 26 at 22:02


















why does the probability have to be to the power 77?
– Toby
Nov 26 at 21:56




why does the probability have to be to the power 77?
– Toby
Nov 26 at 21:56




1




1




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.
– Shubham Johri
Nov 26 at 22:02






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.
– Shubham Johri
Nov 26 at 22:02













0














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





















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
















0














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





















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














0












0








0






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












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 at 21:28









Martin Erhardt

1859




1859












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


















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
















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




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


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3014942%2fwhat-is-the-probability-that-each-element-in-this-string-is-non-zero%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Probability when a professor distributes a quiz and homework assignment to a class of n students.

Image
0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
Read more

Aardman Animations

Image
Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
Read more

Are they similar matrix

Image
0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
Read more