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    a speaks the truth in 70% of cases and b in 90% of the cases. in what percentage of cases are they likely to contradict each other, narrating the same incident?

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    A speaks truth in 70% cases and B in 80% cases. In what percentage of cases they are likely to contradict each other in giving the same statement.

    Click here👆to get an answer to your question ✍️ A speaks truth in 70% cases and B in 80% cases. In what percentage of cases they are likely to contradict each other in giving the same statement.

    Question

    A speaks truth in 70% cases and B in 80% cases. In what percentage of cases they are likely to contradict each other in giving the same statement.

    A

    100 38 ​

    B

    100 62 ​

    C

    100 72 ​

    D

    100 18 ​ Medium Open in App Solution Verified by Toppr

    Correct option is A)

    Let the probability that A and B speak truth be P(A) and P(B) respectively.

    Therefore, P(A)= 100 70 ​ = 10 7 ​ and P(B)= 100 80 ​ = 10 8 ​

    A and B can contradict in stating a fact when one is speaking the truth and other is not speaking the truth.

    Case 1: A is speaking the truth and B is not speaking the truth.

    Required probability =P(A)×(1−P(B))=

    10 7 ​ ×(1− 10 8 ​ )= 100 14 ​

    Case 2: A is not speaking the truth and B is separately the truth.

    Required Probability =(1−P(A))×P(B)=(1−

    10 7 ​ )× 10 8 ​ = 100 24 ​

    Therefore, Percentage of cases in which they are likely to contradict in stating the same fact

    =( 100 14 ​ + 100 24 ​ )=( 100 14+24 ​ )= 100 38 ​

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    [Solved] Atal speaks the truth in 70% of the cases and George speaks

    Concept: P(not A) = 1 - P(A). Probability of a Compound Event [(A and B) or (B and C)] is calculated as: P[(A and B) or (B and C)] = [P(A) × P(B

    Home Mathematics Probability Multiplication Theorem of Events

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    Atal speaks the truth in 70% of the cases and George speaks the truth in 60% cases. In what percentage of cases are they likely to contradict each other in stating the same fact?

    This question was previously asked in

    NIMCET 2013 Official Paper

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    44% 45% 46% 47%

    Answer (Detailed Solution Below)

    Option 3 : 46% Crack with

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    Detailed Solution

    Download Solution PDF

    Concept:

    P(not A) = 1 - P(A).

    Probability of a Compound Event [(A and B) or (B and C)] is calculated as:

    P[(A and B) or (B and C)] = [P(A) × P(B)] + [P(C) × P(D)]

    ('and' means '×' and 'or' means '+')

    Calculation:

    Probability of Atal speaking the truth is 70% = 0.7.

    Probability of Atal NOT speaking the truth is 1 - 0.7 = 0.3.

    Probability of George speaking the truth is 60% = 0.6.

    Probability of George NOT speaking the truth is 1 - 0.6 = 0.4.

    They both will contradict each other in the following case:

    (Atal speaks the truth AND George does not speak the truth) OR (Atal does not speak the truth AND George speaks the truth).

    ∴ The required probability is:

    [P(Atal speaking the truth) × P(George not speaking the truth)] + [P(Atal not speaking the truth) × P(George speaking the truth)]

    = (0.7 × 0.4) + (0.3 × 0.6)

    = 0.28 + 0.18 = 0.46.

    ∴ They both contradict each other in 46% of the cases.

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    स्रोत : testbook.com

    A' speaks truth in 60% cases and 'B' in 70% cases. In what percentages of cases are they likely to contradict each other in stating the same fact?

    A' speaks truth in 60% cases and 'B' in 70% cases. In what percentages of cases are they likely to contradict each other in stating the same fact?

    Home > English > Class 11 > Maths > Chapter > Probability >

    A' speaks truth in 60% cases a...

    A' speaks truth in 60% 60% cases and 'B' in 70% 70%

    cases. In what percentages of cases are they likely to contradict each other in stating the same fact?

    Updated On: 27-06-2022

    00 : 15

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    Text Solution Open Answer in App Answer

    The correct Answer is

    = 18 100 + 28 100 = 46 100 =18100+28100=46100

    Answer

    Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.

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