A die is rolled. Let E be the event “die shows 4” and F be the event “die shows even number”. Are E and F mutually exclusive?​

Question

A die is rolled. Let E be the event “die shows 4” and F be the event “die shows even number”. Are E and F mutually exclusive?​

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Kennedy 1 month 2021-08-16T14:44:08+00:00 2 Answers 0 views 0

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    0
    2021-08-16T14:45:36+00:00

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    \Large\fbox{\color{purple}{QUESTION}}

    A die is rolled. Let E be the event “die shows 4” and F be the event “die shows even number”. Are E and F mutually exclusive?

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    \bf\Huge\red{\mid{\overline{\underline{ ANSWER }}}\mid }

    ➡️Let us assume that 1, 2, 3, 4, 5 and 6 are the possible outcomes when the die is thrown.

    ➡️So, S = (1, 2, 3, 4, 5, 6)

    ➡️As per the conditions given the question

    As per the conditions given the questionE be the event “die shows 4”

    ➡️E = (4)

    ➡️F be the event “die shows even number”

    ➡️F = (2, 4, 6)

    ➡️E∩F = (4) ∩ (2, 4, 6)

    ➡️= 4

    ➡️4 ≠ φ … [because there is a common element in E and F]

    ➡️Therefore E and F are not mutually exclusive event.

    \Large\mathcal\green{FOLLOW \: ME}

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    0
    2021-08-16T14:46:07+00:00

    When a die is rolled the sample space is given by,

    S={1,2,3,4,5,6}

    Now according to question E={4} and F={2,4,6}

    Clearly  E∩F={4}=ϕ

    Hence, E and F are not mutually exclusive events.

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