This is part of a multipart question answered separately on Clay6.com

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

The sample space for two coins tossed one time is $S = (HH, HT, TH, TT)$. The number of total outcomes in 4.

Given E: Tail appears on one coin, $E = (HT, TH).$ The total number of outcomes is 2.

$\Rightarrow P(E) = \large \frac{\text{Number of favorable outcomes in E}}{\text{Total number of outcomes in S}} = \frac{2}{4} = \frac{1}{2}$

Given F: one coin shows head, $F = (HT, TH).$ The total number of outcomes is 2.

$\Rightarrow P(F) = \large \frac{\text{Number of favorable outcomes in F}}{\text{Total number of outcomes in S}} = \frac{2}{4} = \frac{1}{2}$

For our set of events $E \cap F = (HT, TH)$. Total number of outcomes = 2.

$\Rightarrow P(E\;\cap\;F) = \large \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes in S}} = \frac{2}{4} = \frac{1}{2}$

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

$\Rightarrow \large \frac{P(E \;\cap \;F)}{P(F)} $$=\Large \frac {\Large \frac{1}{2}} {\Large \frac{1}{2}}$ = $1$.

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