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Probability

Question
CBSEENMA12033604
Wired Faculty App

A die thrown three times,
E : 4 appears on the third toss,
F : 6 and 5 appears respectively on First two tosses
Determine P(E | F).

Solution
Here a die is thrown three times
 therefore  sample space has 6 x 6 x 6 = 216 outcomes.
therefore space space space straight E space equals space open curly brackets table row cell left parenthesis 1 comma space 1 comma space 4 right parenthesis end cell cell left parenthesis 1 comma space 2 comma space 4 right parenthesis... end cell cell left parenthesis 1 comma space 6 comma space 4 right parenthesis end cell cell left parenthesis 2 comma space 1 comma space 4 right parenthesis end cell cell left parenthesis 2 comma space 2 comma space 4 right parenthesis... end cell cell left parenthesis 2 comma space 6 comma space 4 right parenthesis end cell row cell left parenthesis 3 comma space 1 comma space 4 right parenthesis end cell cell left parenthesis 3 comma space 2 comma space 4 right parenthesis... end cell cell left parenthesis 3 comma space 6 comma space 4 right parenthesis end cell cell left parenthesis 4 comma space 1 comma space 4 right parenthesis end cell cell left parenthesis 4 comma space 2 comma space 4 right parenthesis... end cell cell left parenthesis 4 comma space 6 comma space 4 right parenthesis end cell row cell left parenthesis 5 comma space 1 comma space 4 right parenthesis end cell cell left parenthesis 5 comma space 2 comma space 4 right parenthesis... end cell cell left parenthesis 5 comma space 6 comma space 4 right parenthesis end cell cell left parenthesis 6 comma space 1 comma space 4 right parenthesis end cell cell left parenthesis 6 comma space 2 comma space 4 right parenthesis... end cell cell left parenthesis 6 comma space 6 comma space 4 right parenthesis end cell end table close curly brackets
       F: 6 and 5 appear respectively on first two tosses
therefore space space space space straight F space equals space open curly brackets left parenthesis 6 comma space 5 comma space 1 right parenthesis thin space left parenthesis 6 comma space 5 comma space 2 right parenthesis comma space left parenthesis 6 comma space 5 comma space 3 right parenthesis comma space left parenthesis 6 comma space 5 comma space 4 right parenthesis comma space left parenthesis 6 comma space 5 comma space 5 right parenthesis comma space left parenthesis 6 comma space 5 comma space 6 right parenthesis close curly brackets
      straight E intersection straight F space equals space open curly brackets left parenthesis 6 comma space 5 comma space 4 right parenthesis close curly brackets
Now,  
          straight P left parenthesis straight F right parenthesis space equals space 6 over 216 comma space space straight P left parenthesis straight E intersection straight F right parenthesis space equals space 1 over 216
   therefore 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 space intersection space straight F right parenthesis over denominator straight P left parenthesis straight F right parenthesis end fraction space equals fraction numerator begin display style 1 over 216 end style over denominator begin display style 6 over 216 end style end fraction space equals 1 over 6

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