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Kartik Brar
Asked: 2022-11-09T12:34:47+05:30 In:

A bias dice has probability of getting any of the odd number equal to the inverse of immediate greater even number and the possibility of getting any of the even number are equal, then what is the probability of getting an even number.
1. \(1\over 36\)
2. \(1\over 18\)
3. \(1\over 12\)
4. \(1\over9\)

A bias dice has probability of getting any of the odd number equal to the inverse of immediate greater even number and the possibility of getting any of the even number are equal, then what is the probability of getting an even number.
1. \(1\over 36\)
2. \(1\over 18\)
3. \(1\over 12\)
4. \(1\over9\)
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  1. Correct Answer – Option 3 : \(1\over 12\)

    Concept: 

    General Rule:

    • The number of ways for selecting r from a group of n (n > r) = nCr 
    • The probability of particular case = \(\rm \text{Number of ways for the case can be executed}\over{\text{Total number of ways for selection}}\)
    • The sum of the possible case probabilities = 1

     

    Calculation:

    According to the question;

    Probability of getting ‘1’ = \(1\over2\)

    Probability of getting ‘3’ = \(1\over4\)

    Probability of getting ‘5’ = \(1\over6\)

    Let the probability of getting an even number be x

    As ∑ P = 1

    ⇒ x + \(1\over2\) + \(1\over4\) + \(1\over6\) = 1

    ⇒ x = \(\boldsymbol{1\over12}\)

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