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13.8      A die is thrown three times,

            E : 4 appears on the third toss, F : 6 and 5 appears respectively on first two tosses

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   E : 4 appears on the third toss,         F : 6 and 5 appears respectively on first two tosses

 Total outcomes =6^{3}=216

E=\left \{ 114,124,134,144,154,164,214,224,234,244,254,264,314,324,334,344,354,364,414,424,434,454,464,514,524,534,544,554,564,614,624,634,644,654,664 \right \}  n(E)=36

F=\left \{ 651,652,653,654,655,656 \right \}

n(F)=6

E\cap F=\left \{ 654 \right \}

n(E\cap F)=1

P(E\cap F)=\frac{1}{216}

P( F)=\frac{6}{216}=\frac{1}{36}

P(E| F)=\frac{P(E\cap F)}{P(F)}

P(E| F)=\frac{\frac{1}{216}}{\frac{1}{36}}

P(E| F)=\frac{1}{6}

Posted by

seema garhwal

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