Find the probability that “At least one person…” with multiple variables.












0












$begingroup$


I have a statistics problem and, although I have solved most of it, I am stuck for quite some time in the last question.




Say that there we have a population where the analogy of men to women is θ. Suppose that the probability of a woman being colorblind is $q^2$ and the probability of a man being colorblinf $q$.




Up to now I have found, if choosing one random person, the probability of him/her being colorblind and also that probability of a person being a man, given that he is colorblind. However, now I can't find a way to approach this question:




Suppose we choose two people out of the population radomly. What is the probability that at least one person is colorblind?




I was thinking of approaching it by finding the probability of none being colorblind and then finding its compliment, but I can't figure out how to do this, given that we have a different percentage for men and women. Any help on this would be appreciated.
Thank you!






probability statistics conditional-probability bayes-theorem







share|cite|improve this question









$endgroup$












  • $begingroup$
    To be clear, there are $theta$ men for every $1$ woman, correct?
    $endgroup$
    – Noble Mushtak
    Dec 28 '18 at 22:20






  • 1




    $begingroup$
    No, what I mean is P(A)/ P(A') = θ , where A is that the subject is male. (So I guess, men/women= θ)
    $endgroup$
    – butterflyflyaway
    Dec 28 '18 at 22:22










  • $begingroup$
    Condition the complement of your event on the number of females sampled: 0, 1, or 2.
    $endgroup$
    – Laars Helenius
    Dec 28 '18 at 22:26










  • $begingroup$
    Why only on females though?
    $endgroup$
    – butterflyflyaway
    Dec 28 '18 at 22:27
















0












$begingroup$


I have a statistics problem and, although I have solved most of it, I am stuck for quite some time in the last question.




Say that there we have a population where the analogy of men to women is θ. Suppose that the probability of a woman being colorblind is $q^2$ and the probability of a man being colorblinf $q$.




Up to now I have found, if choosing one random person, the probability of him/her being colorblind and also that probability of a person being a man, given that he is colorblind. However, now I can't find a way to approach this question:




Suppose we choose two people out of the population radomly. What is the probability that at least one person is colorblind?




I was thinking of approaching it by finding the probability of none being colorblind and then finding its compliment, but I can't figure out how to do this, given that we have a different percentage for men and women. Any help on this would be appreciated.
Thank you!






probability statistics conditional-probability bayes-theorem







share|cite|improve this question









$endgroup$












  • $begingroup$
    To be clear, there are $theta$ men for every $1$ woman, correct?
    $endgroup$
    – Noble Mushtak
    Dec 28 '18 at 22:20






  • 1




    $begingroup$
    No, what I mean is P(A)/ P(A') = θ , where A is that the subject is male. (So I guess, men/women= θ)
    $endgroup$
    – butterflyflyaway
    Dec 28 '18 at 22:22










  • $begingroup$
    Condition the complement of your event on the number of females sampled: 0, 1, or 2.
    $endgroup$
    – Laars Helenius
    Dec 28 '18 at 22:26










  • $begingroup$
    Why only on females though?
    $endgroup$
    – butterflyflyaway
    Dec 28 '18 at 22:27














0












0








0





$begingroup$


I have a statistics problem and, although I have solved most of it, I am stuck for quite some time in the last question.




Say that there we have a population where the analogy of men to women is θ. Suppose that the probability of a woman being colorblind is $q^2$ and the probability of a man being colorblinf $q$.




Up to now I have found, if choosing one random person, the probability of him/her being colorblind and also that probability of a person being a man, given that he is colorblind. However, now I can't find a way to approach this question:




Suppose we choose two people out of the population radomly. What is the probability that at least one person is colorblind?




I was thinking of approaching it by finding the probability of none being colorblind and then finding its compliment, but I can't figure out how to do this, given that we have a different percentage for men and women. Any help on this would be appreciated.
Thank you!






probability statistics conditional-probability bayes-theorem







share|cite|improve this question









$endgroup$




I have a statistics problem and, although I have solved most of it, I am stuck for quite some time in the last question.




Say that there we have a population where the analogy of men to women is θ. Suppose that the probability of a woman being colorblind is $q^2$ and the probability of a man being colorblinf $q$.




Up to now I have found, if choosing one random person, the probability of him/her being colorblind and also that probability of a person being a man, given that he is colorblind. However, now I can't find a way to approach this question:




Suppose we choose two people out of the population radomly. What is the probability that at least one person is colorblind?




I was thinking of approaching it by finding the probability of none being colorblind and then finding its compliment, but I can't figure out how to do this, given that we have a different percentage for men and women. Any help on this would be appreciated.
Thank you!






probability statistics conditional-probability bayes-theorem




probability statistics conditional-probability bayes-theorem






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 28 '18 at 22:17









butterflyflyawaybutterflyflyaway

133




133












  • $begingroup$
    To be clear, there are $theta$ men for every $1$ woman, correct?
    $endgroup$
    – Noble Mushtak
    Dec 28 '18 at 22:20






  • 1




    $begingroup$
    No, what I mean is P(A)/ P(A') = θ , where A is that the subject is male. (So I guess, men/women= θ)
    $endgroup$
    – butterflyflyaway
    Dec 28 '18 at 22:22










  • $begingroup$
    Condition the complement of your event on the number of females sampled: 0, 1, or 2.
    $endgroup$
    – Laars Helenius
    Dec 28 '18 at 22:26










  • $begingroup$
    Why only on females though?
    $endgroup$
    – butterflyflyaway
    Dec 28 '18 at 22:27


















  • $begingroup$
    To be clear, there are $theta$ men for every $1$ woman, correct?
    $endgroup$
    – Noble Mushtak
    Dec 28 '18 at 22:20






  • 1




    $begingroup$
    No, what I mean is P(A)/ P(A') = θ , where A is that the subject is male. (So I guess, men/women= θ)
    $endgroup$
    – butterflyflyaway
    Dec 28 '18 at 22:22










  • $begingroup$
    Condition the complement of your event on the number of females sampled: 0, 1, or 2.
    $endgroup$
    – Laars Helenius
    Dec 28 '18 at 22:26










  • $begingroup$
    Why only on females though?
    $endgroup$
    – butterflyflyaway
    Dec 28 '18 at 22:27
















$begingroup$
To be clear, there are $theta$ men for every $1$ woman, correct?
$endgroup$
– Noble Mushtak
Dec 28 '18 at 22:20




$begingroup$
To be clear, there are $theta$ men for every $1$ woman, correct?
$endgroup$
– Noble Mushtak
Dec 28 '18 at 22:20




1




1




$begingroup$
No, what I mean is P(A)/ P(A') = θ , where A is that the subject is male. (So I guess, men/women= θ)
$endgroup$
– butterflyflyaway
Dec 28 '18 at 22:22




$begingroup$
No, what I mean is P(A)/ P(A') = θ , where A is that the subject is male. (So I guess, men/women= θ)
$endgroup$
– butterflyflyaway
Dec 28 '18 at 22:22












$begingroup$
Condition the complement of your event on the number of females sampled: 0, 1, or 2.
$endgroup$
– Laars Helenius
Dec 28 '18 at 22:26




$begingroup$
Condition the complement of your event on the number of females sampled: 0, 1, or 2.
$endgroup$
– Laars Helenius
Dec 28 '18 at 22:26












$begingroup$
Why only on females though?
$endgroup$
– butterflyflyaway
Dec 28 '18 at 22:27




$begingroup$
Why only on females though?
$endgroup$
– butterflyflyaway
Dec 28 '18 at 22:27










1 Answer
1






active

oldest

votes


















2












$begingroup$

Let there be $m$ men and $w$ women (unfortunately, I also have to assume there are also $0$ non-binary people in this population). We have the following:



$$frac m w=thetarightarrow m=theta w$$



Thus, the probabilities of being a man or a woman are:



$$P(man)=frac{m}{m+w}=frac{theta w}{theta w+w}=frac{theta}{theta+1}$$
$$P(woman)=frac{w}{m+w}=frac{w}{theta w+w}=frac{1}{theta+1}$$



Now, we are also given the following:



$$P(colorblind | man)=q^2$$
$$P(colorblind | woman)=q$$



Then, we can use the formula $P(A cap B)=P(B)P(A|B)$ to find the following:



$$P(colorblind cap man)=P(man)P(colorblind | man)=frac{q^2theta}{theta+1}$$
$$P(colorblind cap woman)=P(woman)P(colorblind | woman)=frac{q}{theta+1}$$



Finally, since man and woman are the only two cases in this population, $woman=man^C$—that is, woman is the complement case of man. Thus, we can use the formula $P(A)=P(A cap B)+P(A cap B^C)$ to find:



$$P(colorblind)=P(colorblind cap man)+P(colorblind cap woman)=frac{q^2theta+q}{theta+1}$$



And of course, to find the probability is not colorblind, we use $P(A^C)=1-P(A)$:



$$P(colorblind^C)=1-P(colorblind)=frac{(1-q^2)theta+1-q}{theta+1}$$



Now, the probability that at least one person is colorblind has three cases in it:




  • Both people are colorblind -> $P(colorblind)^2$

  • First person is colorblind, second person is not -> $P(colorblind)cdot P(colorblind^C)$

  • Second person is colorblind, first person is not -> $P(colorblind)cdot P(colorblind^C)$


Thus, to find the total probability of there being one color blind person, we just add up all of these cases:



$$P(at least one colorblind)=P(colorblind)^2+2P(colorblind)cdot P(colorblind^C) \=frac{(q^2theta+q)^2+2(q^2theta+q)(theta(1-q^2)+1-q)}{(theta+1)^2}$$



I'll leave doing the algebra to you. Good luck!






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    To complete the problem, assume the sampling is independent. Then $P(text{at least one colorblind})=1-P(text{not colorblind})^2=1-[1-P(text{colorblind})]^2$
    $endgroup$
    – M. Nestor
    Dec 28 '18 at 22:31












  • $begingroup$
    One more question. This was the answer that I found when I was picking one person and finding the probability of whether he/she is colorblind. So, it remains the same even after I choose two people?
    $endgroup$
    – butterflyflyaway
    Dec 28 '18 at 22:33










  • $begingroup$
    @butterflyflyaway Sorry, see my new edited answer for the actual, final answer of the probability of at least one person being color blind.
    $endgroup$
    – Noble Mushtak
    Dec 28 '18 at 22:36










  • $begingroup$
    @M.Nestor This is a valid solution, but it is equivalent to $1-(1-p)^2=1-(p^2+1-2p)=2p-p^2=p^2+2p-2p^2=p^2+2p(1-p)$, which is what I did.
    $endgroup$
    – Noble Mushtak
    Dec 28 '18 at 22:38












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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Let there be $m$ men and $w$ women (unfortunately, I also have to assume there are also $0$ non-binary people in this population). We have the following:



$$frac m w=thetarightarrow m=theta w$$



Thus, the probabilities of being a man or a woman are:



$$P(man)=frac{m}{m+w}=frac{theta w}{theta w+w}=frac{theta}{theta+1}$$
$$P(woman)=frac{w}{m+w}=frac{w}{theta w+w}=frac{1}{theta+1}$$



Now, we are also given the following:



$$P(colorblind | man)=q^2$$
$$P(colorblind | woman)=q$$



Then, we can use the formula $P(A cap B)=P(B)P(A|B)$ to find the following:



$$P(colorblind cap man)=P(man)P(colorblind | man)=frac{q^2theta}{theta+1}$$
$$P(colorblind cap woman)=P(woman)P(colorblind | woman)=frac{q}{theta+1}$$



Finally, since man and woman are the only two cases in this population, $woman=man^C$—that is, woman is the complement case of man. Thus, we can use the formula $P(A)=P(A cap B)+P(A cap B^C)$ to find:



$$P(colorblind)=P(colorblind cap man)+P(colorblind cap woman)=frac{q^2theta+q}{theta+1}$$



And of course, to find the probability is not colorblind, we use $P(A^C)=1-P(A)$:



$$P(colorblind^C)=1-P(colorblind)=frac{(1-q^2)theta+1-q}{theta+1}$$



Now, the probability that at least one person is colorblind has three cases in it:




  • Both people are colorblind -> $P(colorblind)^2$

  • First person is colorblind, second person is not -> $P(colorblind)cdot P(colorblind^C)$

  • Second person is colorblind, first person is not -> $P(colorblind)cdot P(colorblind^C)$


Thus, to find the total probability of there being one color blind person, we just add up all of these cases:



$$P(at least one colorblind)=P(colorblind)^2+2P(colorblind)cdot P(colorblind^C) \=frac{(q^2theta+q)^2+2(q^2theta+q)(theta(1-q^2)+1-q)}{(theta+1)^2}$$



I'll leave doing the algebra to you. Good luck!






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    To complete the problem, assume the sampling is independent. Then $P(text{at least one colorblind})=1-P(text{not colorblind})^2=1-[1-P(text{colorblind})]^2$
    $endgroup$
    – M. Nestor
    Dec 28 '18 at 22:31












  • $begingroup$
    One more question. This was the answer that I found when I was picking one person and finding the probability of whether he/she is colorblind. So, it remains the same even after I choose two people?
    $endgroup$
    – butterflyflyaway
    Dec 28 '18 at 22:33










  • $begingroup$
    @butterflyflyaway Sorry, see my new edited answer for the actual, final answer of the probability of at least one person being color blind.
    $endgroup$
    – Noble Mushtak
    Dec 28 '18 at 22:36










  • $begingroup$
    @M.Nestor This is a valid solution, but it is equivalent to $1-(1-p)^2=1-(p^2+1-2p)=2p-p^2=p^2+2p-2p^2=p^2+2p(1-p)$, which is what I did.
    $endgroup$
    – Noble Mushtak
    Dec 28 '18 at 22:38
















2












$begingroup$

Let there be $m$ men and $w$ women (unfortunately, I also have to assume there are also $0$ non-binary people in this population). We have the following:



$$frac m w=thetarightarrow m=theta w$$



Thus, the probabilities of being a man or a woman are:



$$P(man)=frac{m}{m+w}=frac{theta w}{theta w+w}=frac{theta}{theta+1}$$
$$P(woman)=frac{w}{m+w}=frac{w}{theta w+w}=frac{1}{theta+1}$$



Now, we are also given the following:



$$P(colorblind | man)=q^2$$
$$P(colorblind | woman)=q$$



Then, we can use the formula $P(A cap B)=P(B)P(A|B)$ to find the following:



$$P(colorblind cap man)=P(man)P(colorblind | man)=frac{q^2theta}{theta+1}$$
$$P(colorblind cap woman)=P(woman)P(colorblind | woman)=frac{q}{theta+1}$$



Finally, since man and woman are the only two cases in this population, $woman=man^C$—that is, woman is the complement case of man. Thus, we can use the formula $P(A)=P(A cap B)+P(A cap B^C)$ to find:



$$P(colorblind)=P(colorblind cap man)+P(colorblind cap woman)=frac{q^2theta+q}{theta+1}$$



And of course, to find the probability is not colorblind, we use $P(A^C)=1-P(A)$:



$$P(colorblind^C)=1-P(colorblind)=frac{(1-q^2)theta+1-q}{theta+1}$$



Now, the probability that at least one person is colorblind has three cases in it:




  • Both people are colorblind -> $P(colorblind)^2$

  • First person is colorblind, second person is not -> $P(colorblind)cdot P(colorblind^C)$

  • Second person is colorblind, first person is not -> $P(colorblind)cdot P(colorblind^C)$


Thus, to find the total probability of there being one color blind person, we just add up all of these cases:



$$P(at least one colorblind)=P(colorblind)^2+2P(colorblind)cdot P(colorblind^C) \=frac{(q^2theta+q)^2+2(q^2theta+q)(theta(1-q^2)+1-q)}{(theta+1)^2}$$



I'll leave doing the algebra to you. Good luck!






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    To complete the problem, assume the sampling is independent. Then $P(text{at least one colorblind})=1-P(text{not colorblind})^2=1-[1-P(text{colorblind})]^2$
    $endgroup$
    – M. Nestor
    Dec 28 '18 at 22:31












  • $begingroup$
    One more question. This was the answer that I found when I was picking one person and finding the probability of whether he/she is colorblind. So, it remains the same even after I choose two people?
    $endgroup$
    – butterflyflyaway
    Dec 28 '18 at 22:33










  • $begingroup$
    @butterflyflyaway Sorry, see my new edited answer for the actual, final answer of the probability of at least one person being color blind.
    $endgroup$
    – Noble Mushtak
    Dec 28 '18 at 22:36










  • $begingroup$
    @M.Nestor This is a valid solution, but it is equivalent to $1-(1-p)^2=1-(p^2+1-2p)=2p-p^2=p^2+2p-2p^2=p^2+2p(1-p)$, which is what I did.
    $endgroup$
    – Noble Mushtak
    Dec 28 '18 at 22:38














2












2








2





$begingroup$

Let there be $m$ men and $w$ women (unfortunately, I also have to assume there are also $0$ non-binary people in this population). We have the following:



$$frac m w=thetarightarrow m=theta w$$



Thus, the probabilities of being a man or a woman are:



$$P(man)=frac{m}{m+w}=frac{theta w}{theta w+w}=frac{theta}{theta+1}$$
$$P(woman)=frac{w}{m+w}=frac{w}{theta w+w}=frac{1}{theta+1}$$



Now, we are also given the following:



$$P(colorblind | man)=q^2$$
$$P(colorblind | woman)=q$$



Then, we can use the formula $P(A cap B)=P(B)P(A|B)$ to find the following:



$$P(colorblind cap man)=P(man)P(colorblind | man)=frac{q^2theta}{theta+1}$$
$$P(colorblind cap woman)=P(woman)P(colorblind | woman)=frac{q}{theta+1}$$



Finally, since man and woman are the only two cases in this population, $woman=man^C$—that is, woman is the complement case of man. Thus, we can use the formula $P(A)=P(A cap B)+P(A cap B^C)$ to find:



$$P(colorblind)=P(colorblind cap man)+P(colorblind cap woman)=frac{q^2theta+q}{theta+1}$$



And of course, to find the probability is not colorblind, we use $P(A^C)=1-P(A)$:



$$P(colorblind^C)=1-P(colorblind)=frac{(1-q^2)theta+1-q}{theta+1}$$



Now, the probability that at least one person is colorblind has three cases in it:




  • Both people are colorblind -> $P(colorblind)^2$

  • First person is colorblind, second person is not -> $P(colorblind)cdot P(colorblind^C)$

  • Second person is colorblind, first person is not -> $P(colorblind)cdot P(colorblind^C)$


Thus, to find the total probability of there being one color blind person, we just add up all of these cases:



$$P(at least one colorblind)=P(colorblind)^2+2P(colorblind)cdot P(colorblind^C) \=frac{(q^2theta+q)^2+2(q^2theta+q)(theta(1-q^2)+1-q)}{(theta+1)^2}$$



I'll leave doing the algebra to you. Good luck!






share|cite|improve this answer











$endgroup$



Let there be $m$ men and $w$ women (unfortunately, I also have to assume there are also $0$ non-binary people in this population). We have the following:



$$frac m w=thetarightarrow m=theta w$$



Thus, the probabilities of being a man or a woman are:



$$P(man)=frac{m}{m+w}=frac{theta w}{theta w+w}=frac{theta}{theta+1}$$
$$P(woman)=frac{w}{m+w}=frac{w}{theta w+w}=frac{1}{theta+1}$$



Now, we are also given the following:



$$P(colorblind | man)=q^2$$
$$P(colorblind | woman)=q$$



Then, we can use the formula $P(A cap B)=P(B)P(A|B)$ to find the following:



$$P(colorblind cap man)=P(man)P(colorblind | man)=frac{q^2theta}{theta+1}$$
$$P(colorblind cap woman)=P(woman)P(colorblind | woman)=frac{q}{theta+1}$$



Finally, since man and woman are the only two cases in this population, $woman=man^C$—that is, woman is the complement case of man. Thus, we can use the formula $P(A)=P(A cap B)+P(A cap B^C)$ to find:



$$P(colorblind)=P(colorblind cap man)+P(colorblind cap woman)=frac{q^2theta+q}{theta+1}$$



And of course, to find the probability is not colorblind, we use $P(A^C)=1-P(A)$:



$$P(colorblind^C)=1-P(colorblind)=frac{(1-q^2)theta+1-q}{theta+1}$$



Now, the probability that at least one person is colorblind has three cases in it:




  • Both people are colorblind -> $P(colorblind)^2$

  • First person is colorblind, second person is not -> $P(colorblind)cdot P(colorblind^C)$

  • Second person is colorblind, first person is not -> $P(colorblind)cdot P(colorblind^C)$


Thus, to find the total probability of there being one color blind person, we just add up all of these cases:



$$P(at least one colorblind)=P(colorblind)^2+2P(colorblind)cdot P(colorblind^C) \=frac{(q^2theta+q)^2+2(q^2theta+q)(theta(1-q^2)+1-q)}{(theta+1)^2}$$



I'll leave doing the algebra to you. Good luck!







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 28 '18 at 22:36

























answered Dec 28 '18 at 22:29









Noble MushtakNoble Mushtak

15.4k1835




15.4k1835








  • 1




    $begingroup$
    To complete the problem, assume the sampling is independent. Then $P(text{at least one colorblind})=1-P(text{not colorblind})^2=1-[1-P(text{colorblind})]^2$
    $endgroup$
    – M. Nestor
    Dec 28 '18 at 22:31












  • $begingroup$
    One more question. This was the answer that I found when I was picking one person and finding the probability of whether he/she is colorblind. So, it remains the same even after I choose two people?
    $endgroup$
    – butterflyflyaway
    Dec 28 '18 at 22:33










  • $begingroup$
    @butterflyflyaway Sorry, see my new edited answer for the actual, final answer of the probability of at least one person being color blind.
    $endgroup$
    – Noble Mushtak
    Dec 28 '18 at 22:36










  • $begingroup$
    @M.Nestor This is a valid solution, but it is equivalent to $1-(1-p)^2=1-(p^2+1-2p)=2p-p^2=p^2+2p-2p^2=p^2+2p(1-p)$, which is what I did.
    $endgroup$
    – Noble Mushtak
    Dec 28 '18 at 22:38














  • 1




    $begingroup$
    To complete the problem, assume the sampling is independent. Then $P(text{at least one colorblind})=1-P(text{not colorblind})^2=1-[1-P(text{colorblind})]^2$
    $endgroup$
    – M. Nestor
    Dec 28 '18 at 22:31












  • $begingroup$
    One more question. This was the answer that I found when I was picking one person and finding the probability of whether he/she is colorblind. So, it remains the same even after I choose two people?
    $endgroup$
    – butterflyflyaway
    Dec 28 '18 at 22:33










  • $begingroup$
    @butterflyflyaway Sorry, see my new edited answer for the actual, final answer of the probability of at least one person being color blind.
    $endgroup$
    – Noble Mushtak
    Dec 28 '18 at 22:36










  • $begingroup$
    @M.Nestor This is a valid solution, but it is equivalent to $1-(1-p)^2=1-(p^2+1-2p)=2p-p^2=p^2+2p-2p^2=p^2+2p(1-p)$, which is what I did.
    $endgroup$
    – Noble Mushtak
    Dec 28 '18 at 22:38








1




1




$begingroup$
To complete the problem, assume the sampling is independent. Then $P(text{at least one colorblind})=1-P(text{not colorblind})^2=1-[1-P(text{colorblind})]^2$
$endgroup$
– M. Nestor
Dec 28 '18 at 22:31






$begingroup$
To complete the problem, assume the sampling is independent. Then $P(text{at least one colorblind})=1-P(text{not colorblind})^2=1-[1-P(text{colorblind})]^2$
$endgroup$
– M. Nestor
Dec 28 '18 at 22:31














$begingroup$
One more question. This was the answer that I found when I was picking one person and finding the probability of whether he/she is colorblind. So, it remains the same even after I choose two people?
$endgroup$
– butterflyflyaway
Dec 28 '18 at 22:33




$begingroup$
One more question. This was the answer that I found when I was picking one person and finding the probability of whether he/she is colorblind. So, it remains the same even after I choose two people?
$endgroup$
– butterflyflyaway
Dec 28 '18 at 22:33












$begingroup$
@butterflyflyaway Sorry, see my new edited answer for the actual, final answer of the probability of at least one person being color blind.
$endgroup$
– Noble Mushtak
Dec 28 '18 at 22:36




$begingroup$
@butterflyflyaway Sorry, see my new edited answer for the actual, final answer of the probability of at least one person being color blind.
$endgroup$
– Noble Mushtak
Dec 28 '18 at 22:36












$begingroup$
@M.Nestor This is a valid solution, but it is equivalent to $1-(1-p)^2=1-(p^2+1-2p)=2p-p^2=p^2+2p-2p^2=p^2+2p(1-p)$, which is what I did.
$endgroup$
– Noble Mushtak
Dec 28 '18 at 22:38




$begingroup$
@M.Nestor This is a valid solution, but it is equivalent to $1-(1-p)^2=1-(p^2+1-2p)=2p-p^2=p^2+2p-2p^2=p^2+2p(1-p)$, which is what I did.
$endgroup$
– Noble Mushtak
Dec 28 '18 at 22:38


















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