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Probability

Question
CBSEENMA12033595

A die is thrown three times. Events A and B are defined as below:
A: 4 on the third throw
B: 6 on the first and 5 on the second throw
Find the probability of A given that B has already occurred.



Solution
The sample space has 216 outcomes.
Now A = 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
       straight B space equals space open curly brackets left parenthesis 6 comma space 5 comma space 1 right parenthesis comma space 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
and straight A intersection straight B 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 B right parenthesis space equals 6 over 216 space and space straight P left parenthesis straight A intersection straight B right parenthesis space equals space 1 over 216
therefore space space space space space straight P left parenthesis straight A space left enclose straight B right parenthesis end enclose space equals space fraction numerator straight P left parenthesis straight A space intersection thin space straight B right parenthesis over denominator straight P left parenthesis straight B right parenthesis end fraction equals space fraction numerator begin display style 1 over 216 end style over denominator begin display style 6 over 216 end style end fraction space equals space 1 over 6
             

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