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

$\begin{array}{1 1} \frac{1}{3} \\ \frac{1}{6} \\ \frac{2}{3} \\ \frac{1}{2} \end{array}$

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A)
• In any problem that involves tossing a coin or a die and determing the outcomes, we can write down the sample space and cout the number of favorable outcomes. Then we find P(E), P(F), P(E$\cap$F) using the set of outcomes.
• Given P(E), P(F), P(E $\cap$ F), P(E/F) $= \large \frac{P(E \;\cap \;F)}{P(F)}$
If a die is thrown three times, the number of elements in the sample space is $6 \times 6 \times 6 = 216$
$E: \begin{matrix} 1,1,4 & 1,2,4 & 1,3,4 & 1,4,4 & 1,5,4 &1,6,4 \\ 2,1,4& .... & & & & \\ 3,1,4 & ... & & & & \\ 4,1,4 & ... & & & & \\ 5,1,4 & ... & & & & \\ 6,1,4&... & & & & 6,6,4 \end{matrix}$
$\Rightarrow P(E) = \large \frac{\text{Number of favorable outcomes in E}}{\text{Total number of outcomes in S}} = \frac{36}{216} = \frac{1}{6}$
$F: (\begin{matrix} 6,5,1 & 6,5,2 & 6,5,3 & 6,5, 4& 6,5,5& 6,5,6 \end{matrix})$. The total number of favorable outcomes = 6.
$\Rightarrow P(F) = \large \frac{\text{Number of favorable outcomes in F}}{\text{Total number of outcomes in S}} = \frac{6}{216} = \frac{1}{36}$
For our set of events $E \cap F =(6,5,4). Total number of outcomes = 1.$\Rightarrow P(E\;\cap\;F) = \large \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes in S}} = \frac{1}{216}$. Given P(E), P(F), P(E$\cap$F), P(E/F)$= \large \frac{P(E \;\cap \;F)}{P(F)}\Rightarrow P(E/F) = \Large\frac {\Large \frac{1}{216}}{\Large\frac{1}{36}}$=$\large\frac{1}{6}\$