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Home » A die is thrown three times, E: 4 appears on the third toss, and 5 appears respectively on first two tosses.
A die is thrown three times, E: 4 appears on the third toss, \mathrm{F}: 6 and 5 appears respectively on first two tosses.
CBSE | Maths | NCERT | Probability
A die is thrown three times, E: 4 appears on the third toss, \mathrm{F}: 6 and 5 appears respectively on first two tosses.

In the sample space, there are 216 outcomes, with each element of the sample space having three entries and taking the form (x, y, z) where 1 \leq x, y, z \leq 6.

Considering the event, E: 4 appears on the third toss

Now the event, F: 6 and 5 appears respectively on the first two tosses

We get the common sample space of both,

\Rightarrow E \cap F={(6,5,4)}

So, P(E)=\frac{36}{216}, P(F)=\frac{6}{216}, P(E \cap F)=\frac{1}{216}

Now, we know that by definition of conditional probability, P(E \mid F)=\frac{P(E \cap F)}{P(F)}

Now by substituting the values we get

\Rightarrow \mathrm{P}(\mathrm{E} \mid \mathrm{F})=\frac{1 / 216}{6 / 216}=\frac{1}{6}

\Rightarrow \mathrm{P}(\mathrm{E} \mid \mathrm{F})=\frac{1}{6}

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