Two coins are tossed once, where
E : no tail appears, F : no head appears
Determine P(E | F).

Solution

Here S = {HH, HT, TH, TT}
E : no tail appears
$therefore space space space straight E space equals space open curly brackets HH close curly brackets$
$straight F colon space no space head space appears$
$therefore space space space space space straight F space equals space open curly brackets TT close curly brackets therefore space space space space straight E intersection straight F space equals straight ϕ therefore space space space space space straight P left parenthesis straight E right parenthesis space equals space 1 fourth comma space space straight P left parenthesis straight F right parenthesis space equals space 1 fourth comma space space space straight P left parenthesis straight E intersection straight F right parenthesis space equals space 0 therefore space space space space space straight P left parenthesis straight E space left enclose straight F right parenthesis space equals space fraction numerator straight P left parenthesis straight E intersection straight F right parenthesis over denominator straight P left parenthesis straight F right parenthesis end fraction space equals fraction numerator 0 over denominator begin display style 1 fourth end style end fraction equals 0$

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Probability

Two coins are tossed once, where
E : no tail appears, F : no head appears
Determine P(E | F).

Some More Questions From Probability Chapter

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Probability

Two coins are tossed once, whereE : no tail appears, F : no head appearsDetermine P(E | F).

Some More Questions From Probability Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Two coins are tossed once, where
E : no tail appears, F : no head appears
Determine P(E | F).