1 Question, 2 Answers












0












$begingroup$


Consider the following problem: In a bullet firing competition $A$ can hit $4$ out of $5$ bullet in target, similar chances of $B$ is $3$ out of $4$ and for $C$ is $2$ out of $3$.They fire simultaneously and exactly two out of them makes correct hit.What is the probability that $C$ has missed the target?
Let $X$: $A$ hits the target, $Y$: $B$ hits the target,$Z$: $C$ hits the target.
Solution $1$: The required probability is ${P(XYZ^c ) over P(XYZ^c )+P(YZX^c )+P(ZXY^c ) }$=${{4 over 5}{3 over 4}{1 over 3} over {4 over 5}{3 over 4}{1 over 3}+{3 over 4}{2 over 3}{1 over 5}+{2 over 3}{4 over 5}{1 over 4}}$=${6 over 13}$.



Solution $2$: The required probability is $P(XYZ^c )={1 over 5}$.



So, my question is which of the solution is correct, and why the other one is incorrect, that is what is the question that the other solution makes that answer$?$






probability-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    The probability that only C misses given that two out of three hit, is the same at the probability that only C misses.
    $endgroup$
    – Math-fun
    Jan 4 at 12:41
















0












$begingroup$


Consider the following problem: In a bullet firing competition $A$ can hit $4$ out of $5$ bullet in target, similar chances of $B$ is $3$ out of $4$ and for $C$ is $2$ out of $3$.They fire simultaneously and exactly two out of them makes correct hit.What is the probability that $C$ has missed the target?
Let $X$: $A$ hits the target, $Y$: $B$ hits the target,$Z$: $C$ hits the target.
Solution $1$: The required probability is ${P(XYZ^c ) over P(XYZ^c )+P(YZX^c )+P(ZXY^c ) }$=${{4 over 5}{3 over 4}{1 over 3} over {4 over 5}{3 over 4}{1 over 3}+{3 over 4}{2 over 3}{1 over 5}+{2 over 3}{4 over 5}{1 over 4}}$=${6 over 13}$.



Solution $2$: The required probability is $P(XYZ^c )={1 over 5}$.



So, my question is which of the solution is correct, and why the other one is incorrect, that is what is the question that the other solution makes that answer$?$






probability-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    The probability that only C misses given that two out of three hit, is the same at the probability that only C misses.
    $endgroup$
    – Math-fun
    Jan 4 at 12:41














0












0








0


1



$begingroup$


Consider the following problem: In a bullet firing competition $A$ can hit $4$ out of $5$ bullet in target, similar chances of $B$ is $3$ out of $4$ and for $C$ is $2$ out of $3$.They fire simultaneously and exactly two out of them makes correct hit.What is the probability that $C$ has missed the target?
Let $X$: $A$ hits the target, $Y$: $B$ hits the target,$Z$: $C$ hits the target.
Solution $1$: The required probability is ${P(XYZ^c ) over P(XYZ^c )+P(YZX^c )+P(ZXY^c ) }$=${{4 over 5}{3 over 4}{1 over 3} over {4 over 5}{3 over 4}{1 over 3}+{3 over 4}{2 over 3}{1 over 5}+{2 over 3}{4 over 5}{1 over 4}}$=${6 over 13}$.



Solution $2$: The required probability is $P(XYZ^c )={1 over 5}$.



So, my question is which of the solution is correct, and why the other one is incorrect, that is what is the question that the other solution makes that answer$?$






probability-theory







share|cite|improve this question









$endgroup$




Consider the following problem: In a bullet firing competition $A$ can hit $4$ out of $5$ bullet in target, similar chances of $B$ is $3$ out of $4$ and for $C$ is $2$ out of $3$.They fire simultaneously and exactly two out of them makes correct hit.What is the probability that $C$ has missed the target?
Let $X$: $A$ hits the target, $Y$: $B$ hits the target,$Z$: $C$ hits the target.
Solution $1$: The required probability is ${P(XYZ^c ) over P(XYZ^c )+P(YZX^c )+P(ZXY^c ) }$=${{4 over 5}{3 over 4}{1 over 3} over {4 over 5}{3 over 4}{1 over 3}+{3 over 4}{2 over 3}{1 over 5}+{2 over 3}{4 over 5}{1 over 4}}$=${6 over 13}$.



Solution $2$: The required probability is $P(XYZ^c )={1 over 5}$.



So, my question is which of the solution is correct, and why the other one is incorrect, that is what is the question that the other solution makes that answer$?$






probability-theory




probability-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 4 at 12:30









Supriyo HalderSupriyo Halder

663113




663113












  • $begingroup$
    The probability that only C misses given that two out of three hit, is the same at the probability that only C misses.
    $endgroup$
    – Math-fun
    Jan 4 at 12:41


















  • $begingroup$
    The probability that only C misses given that two out of three hit, is the same at the probability that only C misses.
    $endgroup$
    – Math-fun
    Jan 4 at 12:41
















$begingroup$
The probability that only C misses given that two out of three hit, is the same at the probability that only C misses.
$endgroup$
– Math-fun
Jan 4 at 12:41




$begingroup$
The probability that only C misses given that two out of three hit, is the same at the probability that only C misses.
$endgroup$
– Math-fun
Jan 4 at 12:41










1 Answer
1






active

oldest

votes


















0












$begingroup$

Solution 1 is correct. The conditional probability (of being given that exactly two of the three participants hits the target) puts you in a restricted sample space. This is why you need to use the conditional probability formula of solution 1.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So what should be the question for the solution 2(to get solution 2 as the answer)?
    $endgroup$
    – Supriyo Halder
    Jan 7 at 12:15










  • $begingroup$
    The question for which 'solution 2' is the answer would be "what is the probability that A and B both hit the target and C misses the target?".
    $endgroup$
    – paw88789
    Jan 7 at 16:27










  • $begingroup$
    But, A and B both hit the target and C misses the target is same as exactly two of them hit the target and C missed the target.
    $endgroup$
    – Supriyo Halder
    Jan 8 at 12:18










  • $begingroup$
    Yes, but in the actual problem, you are given the additional information that exactly two of the three hit the target. This additional information causes you to have a reduced sample space. That is we're interested in ABC' occurring but only out of the possibilities ABC', AB'C, A'BC. All the other possibilities are no longer relevant.
    $endgroup$
    – paw88789
    Jan 8 at 13:08











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1 Answer
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1 Answer
1






active

oldest

votes









active

oldest

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active

oldest

votes









0












$begingroup$

Solution 1 is correct. The conditional probability (of being given that exactly two of the three participants hits the target) puts you in a restricted sample space. This is why you need to use the conditional probability formula of solution 1.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So what should be the question for the solution 2(to get solution 2 as the answer)?
    $endgroup$
    – Supriyo Halder
    Jan 7 at 12:15










  • $begingroup$
    The question for which 'solution 2' is the answer would be "what is the probability that A and B both hit the target and C misses the target?".
    $endgroup$
    – paw88789
    Jan 7 at 16:27










  • $begingroup$
    But, A and B both hit the target and C misses the target is same as exactly two of them hit the target and C missed the target.
    $endgroup$
    – Supriyo Halder
    Jan 8 at 12:18










  • $begingroup$
    Yes, but in the actual problem, you are given the additional information that exactly two of the three hit the target. This additional information causes you to have a reduced sample space. That is we're interested in ABC' occurring but only out of the possibilities ABC', AB'C, A'BC. All the other possibilities are no longer relevant.
    $endgroup$
    – paw88789
    Jan 8 at 13:08
















0












$begingroup$

Solution 1 is correct. The conditional probability (of being given that exactly two of the three participants hits the target) puts you in a restricted sample space. This is why you need to use the conditional probability formula of solution 1.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So what should be the question for the solution 2(to get solution 2 as the answer)?
    $endgroup$
    – Supriyo Halder
    Jan 7 at 12:15










  • $begingroup$
    The question for which 'solution 2' is the answer would be "what is the probability that A and B both hit the target and C misses the target?".
    $endgroup$
    – paw88789
    Jan 7 at 16:27










  • $begingroup$
    But, A and B both hit the target and C misses the target is same as exactly two of them hit the target and C missed the target.
    $endgroup$
    – Supriyo Halder
    Jan 8 at 12:18










  • $begingroup$
    Yes, but in the actual problem, you are given the additional information that exactly two of the three hit the target. This additional information causes you to have a reduced sample space. That is we're interested in ABC' occurring but only out of the possibilities ABC', AB'C, A'BC. All the other possibilities are no longer relevant.
    $endgroup$
    – paw88789
    Jan 8 at 13:08














0












0








0





$begingroup$

Solution 1 is correct. The conditional probability (of being given that exactly two of the three participants hits the target) puts you in a restricted sample space. This is why you need to use the conditional probability formula of solution 1.






share|cite|improve this answer









$endgroup$



Solution 1 is correct. The conditional probability (of being given that exactly two of the three participants hits the target) puts you in a restricted sample space. This is why you need to use the conditional probability formula of solution 1.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 4 at 12:48









paw88789paw88789

29.4k12349




29.4k12349












  • $begingroup$
    So what should be the question for the solution 2(to get solution 2 as the answer)?
    $endgroup$
    – Supriyo Halder
    Jan 7 at 12:15










  • $begingroup$
    The question for which 'solution 2' is the answer would be "what is the probability that A and B both hit the target and C misses the target?".
    $endgroup$
    – paw88789
    Jan 7 at 16:27










  • $begingroup$
    But, A and B both hit the target and C misses the target is same as exactly two of them hit the target and C missed the target.
    $endgroup$
    – Supriyo Halder
    Jan 8 at 12:18










  • $begingroup$
    Yes, but in the actual problem, you are given the additional information that exactly two of the three hit the target. This additional information causes you to have a reduced sample space. That is we're interested in ABC' occurring but only out of the possibilities ABC', AB'C, A'BC. All the other possibilities are no longer relevant.
    $endgroup$
    – paw88789
    Jan 8 at 13:08


















  • $begingroup$
    So what should be the question for the solution 2(to get solution 2 as the answer)?
    $endgroup$
    – Supriyo Halder
    Jan 7 at 12:15










  • $begingroup$
    The question for which 'solution 2' is the answer would be "what is the probability that A and B both hit the target and C misses the target?".
    $endgroup$
    – paw88789
    Jan 7 at 16:27










  • $begingroup$
    But, A and B both hit the target and C misses the target is same as exactly two of them hit the target and C missed the target.
    $endgroup$
    – Supriyo Halder
    Jan 8 at 12:18










  • $begingroup$
    Yes, but in the actual problem, you are given the additional information that exactly two of the three hit the target. This additional information causes you to have a reduced sample space. That is we're interested in ABC' occurring but only out of the possibilities ABC', AB'C, A'BC. All the other possibilities are no longer relevant.
    $endgroup$
    – paw88789
    Jan 8 at 13:08
















$begingroup$
So what should be the question for the solution 2(to get solution 2 as the answer)?
$endgroup$
– Supriyo Halder
Jan 7 at 12:15




$begingroup$
So what should be the question for the solution 2(to get solution 2 as the answer)?
$endgroup$
– Supriyo Halder
Jan 7 at 12:15












$begingroup$
The question for which 'solution 2' is the answer would be "what is the probability that A and B both hit the target and C misses the target?".
$endgroup$
– paw88789
Jan 7 at 16:27




$begingroup$
The question for which 'solution 2' is the answer would be "what is the probability that A and B both hit the target and C misses the target?".
$endgroup$
– paw88789
Jan 7 at 16:27












$begingroup$
But, A and B both hit the target and C misses the target is same as exactly two of them hit the target and C missed the target.
$endgroup$
– Supriyo Halder
Jan 8 at 12:18




$begingroup$
But, A and B both hit the target and C misses the target is same as exactly two of them hit the target and C missed the target.
$endgroup$
– Supriyo Halder
Jan 8 at 12:18












$begingroup$
Yes, but in the actual problem, you are given the additional information that exactly two of the three hit the target. This additional information causes you to have a reduced sample space. That is we're interested in ABC' occurring but only out of the possibilities ABC', AB'C, A'BC. All the other possibilities are no longer relevant.
$endgroup$
– paw88789
Jan 8 at 13:08




$begingroup$
Yes, but in the actual problem, you are given the additional information that exactly two of the three hit the target. This additional information causes you to have a reduced sample space. That is we're interested in ABC' occurring but only out of the possibilities ABC', AB'C, A'BC. All the other possibilities are no longer relevant.
$endgroup$
– paw88789
Jan 8 at 13:08


















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