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# Determine P(E|F) when Two coins are tossed once, where E : no tail appears, F : no head appears

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

Toolbox:
• 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: No tail appears, $E = (HH).$ The total number of outcomes is 1.
$\Rightarrow P(E) = \large \frac{\text{Number of favorable outcomes in E}}{\text{Total number of outcomes in S}} = \frac{1}{4}$
Given F: No head appears. $F = (TT).$ The total number of outcomes is 1.
$\Rightarrow P(F) = \large \frac{\text{Number of favorable outcomes in F}}{\text{Total number of outcomes in S}} = \frac{1}{4}$
For our set of events $E \cap F =$ Empty Set. Total number of outcomes = 0.
$\Rightarrow P(E\;\cap\;F) = \large \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes in S}}$ = 0.
Given P(E), P(F), P(E $\cap$ F), P(E/F) $= \large \frac{P(E \;\cap \;F)}{P(F)}$
$\Rightarrow \large \frac{P(E \;\cap \;F)}{P(F)}$= 0.