A and B likely to contradict each other in stating same fact?
$begingroup$
A speaks truth in $75%$ cases and B in $80%$ cases. In what percentage of cases are they likely to contradict each other in stating the same fact?
(a) $70%$ (b)$35%$
(c) $25%$ (d)$20%$
what i have tried:
I would like to show about what i think, If QA is in Yes/No both A and B answer behave in similar to Exclusive-OR where agree$=0$ and disagree$=1$
$$begin{array}{l|l|l}
text{A} & text{B} & text{Fact} \ hline
T & T & text{agree} \ hline
T & F & text{disagree} \ hline
F & T & text{disagree} \ hline
F & F & text{agree}
end{array}$$
Percentage of cases where they are likely to contradict each other=$80%-75%=5%$
because till $75%$ A speaks the truth and even B speaks truth
After $75%$ A speaks false and B speaks truth and after $80%$ both A & B speaks false
and so the answer is $5%$ which is not there in option can any please help me this problem?
probability permutations combinations percentages
asked Feb 5 '15 at 9:07
CY5CY5
160113
$endgroup$
add a comment |
$begingroup$
A speaks truth in $75%$ cases and B in $80%$ cases. In what percentage of cases are they likely to contradict each other in stating the same fact?
(a) $70%$ (b)$35%$
(c) $25%$ (d)$20%$
what i have tried:
I would like to show about what i think, If QA is in Yes/No both A and B answer behave in similar to Exclusive-OR where agree$=0$ and disagree$=1$
$$begin{array}{l|l|l}
text{A} & text{B} & text{Fact} \ hline
T & T & text{agree} \ hline
T & F & text{disagree} \ hline
F & T & text{disagree} \ hline
F & F & text{agree}
end{array}$$
Percentage of cases where they are likely to contradict each other=$80%-75%=5%$
because till $75%$ A speaks the truth and even B speaks truth
After $75%$ A speaks false and B speaks truth and after $80%$ both A & B speaks false
and so the answer is $5%$ which is not there in option can any please help me this problem?
probability permutations combinations percentages
asked Feb 5 '15 at 9:07
CY5CY5
160113
$endgroup$
add a comment |
$begingroup$
A speaks truth in $75%$ cases and B in $80%$ cases. In what percentage of cases are they likely to contradict each other in stating the same fact?
(a) $70%$ (b)$35%$
(c) $25%$ (d)$20%$
what i have tried:
I would like to show about what i think, If QA is in Yes/No both A and B answer behave in similar to Exclusive-OR where agree$=0$ and disagree$=1$
$$begin{array}{l|l|l}
text{A} & text{B} & text{Fact} \ hline
T & T & text{agree} \ hline
T & F & text{disagree} \ hline
F & T & text{disagree} \ hline
F & F & text{agree}
end{array}$$
Percentage of cases where they are likely to contradict each other=$80%-75%=5%$
because till $75%$ A speaks the truth and even B speaks truth
After $75%$ A speaks false and B speaks truth and after $80%$ both A & B speaks false
and so the answer is $5%$ which is not there in option can any please help me this problem?
probability permutations combinations percentages
asked Feb 5 '15 at 9:07
CY5CY5
160113
$endgroup$
A speaks truth in $75%$ cases and B in $80%$ cases. In what percentage of cases are they likely to contradict each other in stating the same fact?
(a) $70%$ (b)$35%$
(c) $25%$ (d)$20%$
what i have tried:
I would like to show about what i think, If QA is in Yes/No both A and B answer behave in similar to Exclusive-OR where agree$=0$ and disagree$=1$
$$begin{array}{l|l|l}
text{A} & text{B} & text{Fact} \ hline
T & T & text{agree} \ hline
T & F & text{disagree} \ hline
F & T & text{disagree} \ hline
F & F & text{agree}
end{array}$$
Percentage of cases where they are likely to contradict each other=$80%-75%=5%$
because till $75%$ A speaks the truth and even B speaks truth
After $75%$ A speaks false and B speaks truth and after $80%$ both A & B speaks false
and so the answer is $5%$ which is not there in option can any please help me this problem?
probability permutations combinations percentages
probability permutations combinations percentages
asked Feb 5 '15 at 9:07
CY5CY5
160113
asked Feb 5 '15 at 9:07
CY5CY5
160113
asked Feb 5 '15 at 9:07
CY5CY5
160113
asked Feb 5 '15 at 9:07
CY5CY5
160113
asked Feb 5 '15 at 9:07
CY5CY5
160113
160113
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The probability that they contradict each other is given by (TF, FT) as you mentioned in the table. Assuming both the speakers choose to speak truth or false independently, the probability can be calculated as follows-
Suppose the probability A and B speak truth is given by $P_A$, $P_B$ respectively. Their probability of speaking false is given by $1-P_A$ and $1-P_B$, as speaking truth and false are mutually exclusive. The probability that they contradict each other is given by $P_A(1-P_B)+(1-P_A)P_B$
answered Feb 5 '15 at 9:38
CuriousCurious
889516
$endgroup$
add a comment |
$begingroup$
Since it is not stated that they decide independantly whether or not to tell the truth, anything between $5%$ and $45%$ is possible.
answered Feb 5 '15 at 9:12
Hagen von EitzenHagen von Eitzen
279k23271502
$endgroup$
$begingroup$
how did you get $45%$?
$endgroup$
– CY5
Feb 5 '15 at 9:35
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The probability that they contradict each other is given by (TF, FT) as you mentioned in the table. Assuming both the speakers choose to speak truth or false independently, the probability can be calculated as follows-
Suppose the probability A and B speak truth is given by $P_A$, $P_B$ respectively. Their probability of speaking false is given by $1-P_A$ and $1-P_B$, as speaking truth and false are mutually exclusive. The probability that they contradict each other is given by $P_A(1-P_B)+(1-P_A)P_B$
answered Feb 5 '15 at 9:38
CuriousCurious
889516
$endgroup$
add a comment |
$begingroup$
The probability that they contradict each other is given by (TF, FT) as you mentioned in the table. Assuming both the speakers choose to speak truth or false independently, the probability can be calculated as follows-
Suppose the probability A and B speak truth is given by $P_A$, $P_B$ respectively. Their probability of speaking false is given by $1-P_A$ and $1-P_B$, as speaking truth and false are mutually exclusive. The probability that they contradict each other is given by $P_A(1-P_B)+(1-P_A)P_B$
answered Feb 5 '15 at 9:38
CuriousCurious
889516
$endgroup$
add a comment |
$begingroup$
The probability that they contradict each other is given by (TF, FT) as you mentioned in the table. Assuming both the speakers choose to speak truth or false independently, the probability can be calculated as follows-
Suppose the probability A and B speak truth is given by $P_A$, $P_B$ respectively. Their probability of speaking false is given by $1-P_A$ and $1-P_B$, as speaking truth and false are mutually exclusive. The probability that they contradict each other is given by $P_A(1-P_B)+(1-P_A)P_B$
answered Feb 5 '15 at 9:38
CuriousCurious
889516
$endgroup$
The probability that they contradict each other is given by (TF, FT) as you mentioned in the table. Assuming both the speakers choose to speak truth or false independently, the probability can be calculated as follows-
Suppose the probability A and B speak truth is given by $P_A$, $P_B$ respectively. Their probability of speaking false is given by $1-P_A$ and $1-P_B$, as speaking truth and false are mutually exclusive. The probability that they contradict each other is given by $P_A(1-P_B)+(1-P_A)P_B$
answered Feb 5 '15 at 9:38
CuriousCurious
889516
answered Feb 5 '15 at 9:38
CuriousCurious
889516
answered Feb 5 '15 at 9:38
CuriousCurious
889516
answered Feb 5 '15 at 9:38
CuriousCurious
889516
889516
add a comment |
add a comment |
$begingroup$
Since it is not stated that they decide independantly whether or not to tell the truth, anything between $5%$ and $45%$ is possible.
answered Feb 5 '15 at 9:12
Hagen von EitzenHagen von Eitzen
279k23271502
$endgroup$
$begingroup$
how did you get $45%$?
$endgroup$
– CY5
Feb 5 '15 at 9:35
add a comment |
$begingroup$
Since it is not stated that they decide independantly whether or not to tell the truth, anything between $5%$ and $45%$ is possible.
answered Feb 5 '15 at 9:12
Hagen von EitzenHagen von Eitzen
279k23271502
$endgroup$
$begingroup$
how did you get $45%$?
$endgroup$
– CY5
Feb 5 '15 at 9:35
add a comment |
$begingroup$
Since it is not stated that they decide independantly whether or not to tell the truth, anything between $5%$ and $45%$ is possible.
answered Feb 5 '15 at 9:12
Hagen von EitzenHagen von Eitzen
279k23271502
$endgroup$
Since it is not stated that they decide independantly whether or not to tell the truth, anything between $5%$ and $45%$ is possible.
answered Feb 5 '15 at 9:12
Hagen von EitzenHagen von Eitzen
279k23271502
answered Feb 5 '15 at 9:12
Hagen von EitzenHagen von Eitzen
279k23271502
answered Feb 5 '15 at 9:12
Hagen von EitzenHagen von Eitzen
279k23271502
answered Feb 5 '15 at 9:12
Hagen von EitzenHagen von Eitzen
279k23271502
279k23271502
$begingroup$
how did you get $45%$?
$endgroup$
– CY5
Feb 5 '15 at 9:35
add a comment |
$begingroup$
how did you get $45%$?
$endgroup$
– CY5
Feb 5 '15 at 9:35
$begingroup$
how did you get $45%$?
$endgroup$
– CY5
Feb 5 '15 at 9:35
$begingroup$
how did you get $45%$?
$endgroup$
– CY5
Feb 5 '15 at 9:35
add a comment |
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