$\begin{array}{1 1} 0 \\ 1 \\ \frac{1}{2} \\ \frac{1}{3} \end{array} $

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- In any problem that involves 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) $= \large \frac{P(E \;\cap \;F)}{P(F)}$

Given the mother M, father F and son S line up for a family picture, the sample space will be S: (MFS, MSF, FMS, FSM, SMF, SFM) and the total number of outcomes = 6.

For the event E: where the son is on one end, E = (MFS, FMS, SMF, SFM). The total number of outcomes = 4.

$\Rightarrow P(E) = \large \frac{\text{Number of favorable outcomes in E}}{\text{Total number of outcomes in S}} = \frac{4}{6} = \frac{2}{3}$

For the event F: where the father is in the middle, F = (MFS, SFM). The total number of outcomes = 2.

$\Rightarrow P(F) = \large \frac{\text{Number of favorable outcomes in F}}{\text{Total number of outcomes in S}} = \frac{2}{6} = \frac{1}{3}$

For our events, E $\cap$ F = (MFS, SFM). The total number of outcomes = 2.

$\Rightarrow P(E \cap F) = \large \frac{\text{Number of favorable outcomes in E $\cap$ F}}{\text{Total number of outcomes in S}} = \frac{2}{6} = \frac{1}{3}$

Given P(E), P(F), P(E $\cap$ F), P(E/F) $= \large \frac{P(E \;\cap \;F)}{P(F)}$

$\Rightarrow P(E/F) = \Large\frac {\Large \frac{1}{3}}{\Large\frac{1}{3}}$ = 1.

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