Mixing
How to find out the probability to enter a house?
$begingroup$
Let's assume there is $50%$ of chances to someone's house entry door to be locked. A locksmith has $10$ keys, of which only 2 open the door. What is the probability of someone to enter the house through this door if he (or she) can choose by chance only one key among the ten?
The question also asks to provide the sample space related to this random experiment.
MY ATTEMPT
I have obtained the following sample space
begin{align*}
Omega := & {(K_{1},L^{c}),(K_{2},L^{c}),(K_{3},L^{c}),(K_{4},L^{c}),(K_{5},L^{c}),(K_{6},L^{c}),(K_{7},L^{c}),(K_{8},L^{c}),\
& (K_{9},L^{c}),(K_{10},L^{c}),(K_{1},L),(K_{2},L),(K_{3},L),(K_{4},L),(K_{5},L),(K_{6},L),(K_{7},L),\
&(K_{8},L),(K_{9},L),(K_{10},L)}
end{align*}
Where $K_{i}$ indicates the "choice of the $i$-th key" (for $1leq i leq 10$), $L^{c}$ indicates the door is open and $L$ indicates the door is locked. Since there are two keys which open the door, there are twelve cases in which the person can enter the house.
Due to the fact that the outcomes have the same weight, the sought probability is given by
begin{align*}
mathbb{P}(E) = frac{12}{20} = 0.6
end{align*}
Where $E$ indicates the event "the person has entered the house". Here I have used the fact that the events $K_{i}$ and $L$ are independent.
Hence I'd like to ask: am I on the right track? Is there another way to obtain the right result? Could someone justify/rectify my argument? Any help is appreciated. Thanks in advance.
probability probability-theory
asked Jan 6 at 23:41
user1337user1337
42110
$endgroup$
add a comment |
$begingroup$
Let's assume there is $50%$ of chances to someone's house entry door to be locked. A locksmith has $10$ keys, of which only 2 open the door. What is the probability of someone to enter the house through this door if he (or she) can choose by chance only one key among the ten?
The question also asks to provide the sample space related to this random experiment.
MY ATTEMPT
I have obtained the following sample space
begin{align*}
Omega := & {(K_{1},L^{c}),(K_{2},L^{c}),(K_{3},L^{c}),(K_{4},L^{c}),(K_{5},L^{c}),(K_{6},L^{c}),(K_{7},L^{c}),(K_{8},L^{c}),\
& (K_{9},L^{c}),(K_{10},L^{c}),(K_{1},L),(K_{2},L),(K_{3},L),(K_{4},L),(K_{5},L),(K_{6},L),(K_{7},L),\
&(K_{8},L),(K_{9},L),(K_{10},L)}
end{align*}
Where $K_{i}$ indicates the "choice of the $i$-th key" (for $1leq i leq 10$), $L^{c}$ indicates the door is open and $L$ indicates the door is locked. Since there are two keys which open the door, there are twelve cases in which the person can enter the house.
Due to the fact that the outcomes have the same weight, the sought probability is given by
begin{align*}
mathbb{P}(E) = frac{12}{20} = 0.6
end{align*}
Where $E$ indicates the event "the person has entered the house". Here I have used the fact that the events $K_{i}$ and $L$ are independent.
Hence I'd like to ask: am I on the right track? Is there another way to obtain the right result? Could someone justify/rectify my argument? Any help is appreciated. Thanks in advance.
probability probability-theory
asked Jan 6 at 23:41
user1337user1337
42110
$endgroup$
7
$begingroup$
Answer is correct. I would solve the problem the following way: $P(E)=P($house is unlocked$) + P($house is locked$)cdot P($right key is chosen$) = 1/2 + 1/2cdot 2/10 = 6/10=0.6$.
$endgroup$
– EuxhenH
Jan 6 at 23:55
$begingroup$
@EuxhenH (+1) I think you should put that in an answer instead of a comment.
$endgroup$
– Thomas Bladt
Jan 7 at 1:11
add a comment |
$begingroup$
Let's assume there is $50%$ of chances to someone's house entry door to be locked. A locksmith has $10$ keys, of which only 2 open the door. What is the probability of someone to enter the house through this door if he (or she) can choose by chance only one key among the ten?
The question also asks to provide the sample space related to this random experiment.
MY ATTEMPT
I have obtained the following sample space
begin{align*}
Omega := & {(K_{1},L^{c}),(K_{2},L^{c}),(K_{3},L^{c}),(K_{4},L^{c}),(K_{5},L^{c}),(K_{6},L^{c}),(K_{7},L^{c}),(K_{8},L^{c}),\
& (K_{9},L^{c}),(K_{10},L^{c}),(K_{1},L),(K_{2},L),(K_{3},L),(K_{4},L),(K_{5},L),(K_{6},L),(K_{7},L),\
&(K_{8},L),(K_{9},L),(K_{10},L)}
end{align*}
Where $K_{i}$ indicates the "choice of the $i$-th key" (for $1leq i leq 10$), $L^{c}$ indicates the door is open and $L$ indicates the door is locked. Since there are two keys which open the door, there are twelve cases in which the person can enter the house.
Due to the fact that the outcomes have the same weight, the sought probability is given by
begin{align*}
mathbb{P}(E) = frac{12}{20} = 0.6
end{align*}
Where $E$ indicates the event "the person has entered the house". Here I have used the fact that the events $K_{i}$ and $L$ are independent.
Hence I'd like to ask: am I on the right track? Is there another way to obtain the right result? Could someone justify/rectify my argument? Any help is appreciated. Thanks in advance.
probability probability-theory
asked Jan 6 at 23:41
user1337user1337
42110
$endgroup$
Let's assume there is $50%$ of chances to someone's house entry door to be locked. A locksmith has $10$ keys, of which only 2 open the door. What is the probability of someone to enter the house through this door if he (or she) can choose by chance only one key among the ten?
The question also asks to provide the sample space related to this random experiment.
MY ATTEMPT
I have obtained the following sample space
begin{align*}
Omega := & {(K_{1},L^{c}),(K_{2},L^{c}),(K_{3},L^{c}),(K_{4},L^{c}),(K_{5},L^{c}),(K_{6},L^{c}),(K_{7},L^{c}),(K_{8},L^{c}),\
& (K_{9},L^{c}),(K_{10},L^{c}),(K_{1},L),(K_{2},L),(K_{3},L),(K_{4},L),(K_{5},L),(K_{6},L),(K_{7},L),\
&(K_{8},L),(K_{9},L),(K_{10},L)}
end{align*}
Where $K_{i}$ indicates the "choice of the $i$-th key" (for $1leq i leq 10$), $L^{c}$ indicates the door is open and $L$ indicates the door is locked. Since there are two keys which open the door, there are twelve cases in which the person can enter the house.
Due to the fact that the outcomes have the same weight, the sought probability is given by
begin{align*}
mathbb{P}(E) = frac{12}{20} = 0.6
end{align*}
Where $E$ indicates the event "the person has entered the house". Here I have used the fact that the events $K_{i}$ and $L$ are independent.
Hence I'd like to ask: am I on the right track? Is there another way to obtain the right result? Could someone justify/rectify my argument? Any help is appreciated. Thanks in advance.
probability probability-theory
probability probability-theory
asked Jan 6 at 23:41
user1337user1337
42110
asked Jan 6 at 23:41
user1337user1337
42110
edited Jan 6 at 23:53
user1337
asked Jan 6 at 23:41
user1337user1337
42110
asked Jan 6 at 23:41
user1337user1337
42110
asked Jan 6 at 23:41
user1337user1337
42110
42110
7
$begingroup$
Answer is correct. I would solve the problem the following way: $P(E)=P($house is unlocked$) + P($house is locked$)cdot P($right key is chosen$) = 1/2 + 1/2cdot 2/10 = 6/10=0.6$.
$endgroup$
– EuxhenH
Jan 6 at 23:55
$begingroup$
@EuxhenH (+1) I think you should put that in an answer instead of a comment.
$endgroup$
– Thomas Bladt
Jan 7 at 1:11
add a comment |
7
$begingroup$
Answer is correct. I would solve the problem the following way: $P(E)=P($house is unlocked$) + P($house is locked$)cdot P($right key is chosen$) = 1/2 + 1/2cdot 2/10 = 6/10=0.6$.
$endgroup$
– EuxhenH
Jan 6 at 23:55
$begingroup$
@EuxhenH (+1) I think you should put that in an answer instead of a comment.
$endgroup$
– Thomas Bladt
Jan 7 at 1:11
7
7
$begingroup$
Answer is correct. I would solve the problem the following way: $P(E)=P($house is unlocked$) + P($house is locked$)cdot P($right key is chosen$) = 1/2 + 1/2cdot 2/10 = 6/10=0.6$.
$endgroup$
– EuxhenH
Jan 6 at 23:55
$begingroup$
Answer is correct. I would solve the problem the following way: $P(E)=P($house is unlocked$) + P($house is locked$)cdot P($right key is chosen$) = 1/2 + 1/2cdot 2/10 = 6/10=0.6$.
$endgroup$
– EuxhenH
Jan 6 at 23:55
$begingroup$
@EuxhenH (+1) I think you should put that in an answer instead of a comment.
$endgroup$
– Thomas Bladt
Jan 7 at 1:11
$begingroup$
@EuxhenH (+1) I think you should put that in an answer instead of a comment.
$endgroup$
– Thomas Bladt
Jan 7 at 1:11
add a comment |
1 Answer
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$begingroup$
The answer is correct although there is a more systematical approach to the problem in hand. Someone entering the house means that either the house is unlocked or the house is locked and the person chooses the right key. In terms of probability this means
$$begin{array}{crl}P(E) &=& P(house is unlocked) + P(house is locked)cdot P(right key is chosen)\ &=& dfrac{1}{2}+dfrac{1}{2}cdot dfrac{2}{10}\&=& dfrac{6}{10}end{array}$$
answered Jan 7 at 6:53
EuxhenHEuxhenH
484210
$endgroup$
add a comment |
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1 Answer
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$begingroup$
The answer is correct although there is a more systematical approach to the problem in hand. Someone entering the house means that either the house is unlocked or the house is locked and the person chooses the right key. In terms of probability this means
$$begin{array}{crl}P(E) &=& P(house is unlocked) + P(house is locked)cdot P(right key is chosen)\ &=& dfrac{1}{2}+dfrac{1}{2}cdot dfrac{2}{10}\&=& dfrac{6}{10}end{array}$$
answered Jan 7 at 6:53
EuxhenHEuxhenH
484210
$endgroup$
add a comment |
$begingroup$
The answer is correct although there is a more systematical approach to the problem in hand. Someone entering the house means that either the house is unlocked or the house is locked and the person chooses the right key. In terms of probability this means
$$begin{array}{crl}P(E) &=& P(house is unlocked) + P(house is locked)cdot P(right key is chosen)\ &=& dfrac{1}{2}+dfrac{1}{2}cdot dfrac{2}{10}\&=& dfrac{6}{10}end{array}$$
answered Jan 7 at 6:53
EuxhenHEuxhenH
484210
$endgroup$
add a comment |
$begingroup$
The answer is correct although there is a more systematical approach to the problem in hand. Someone entering the house means that either the house is unlocked or the house is locked and the person chooses the right key. In terms of probability this means
$$begin{array}{crl}P(E) &=& P(house is unlocked) + P(house is locked)cdot P(right key is chosen)\ &=& dfrac{1}{2}+dfrac{1}{2}cdot dfrac{2}{10}\&=& dfrac{6}{10}end{array}$$
answered Jan 7 at 6:53
EuxhenHEuxhenH
484210
$endgroup$
The answer is correct although there is a more systematical approach to the problem in hand. Someone entering the house means that either the house is unlocked or the house is locked and the person chooses the right key. In terms of probability this means
$$begin{array}{crl}P(E) &=& P(house is unlocked) + P(house is locked)cdot P(right key is chosen)\ &=& dfrac{1}{2}+dfrac{1}{2}cdot dfrac{2}{10}\&=& dfrac{6}{10}end{array}$$
answered Jan 7 at 6:53
EuxhenHEuxhenH
484210
answered Jan 7 at 6:53
EuxhenHEuxhenH
484210
answered Jan 7 at 6:53
EuxhenHEuxhenH
484210
answered Jan 7 at 6:53
EuxhenHEuxhenH
484210
484210
add a comment |
add a comment |
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7
$begingroup$
Answer is correct. I would solve the problem the following way: $P(E)=P($house is unlocked$) + P($house is locked$)cdot P($right key is chosen$) = 1/2 + 1/2cdot 2/10 = 6/10=0.6$.
$endgroup$
– EuxhenH
Jan 6 at 23:55
$begingroup$
@EuxhenH (+1) I think you should put that in an answer instead of a comment.
$endgroup$
– Thomas Bladt
Jan 7 at 1:11