# Determine P(E|F) when a coin is tossed three times, where E : at most two tails , F : at least one tail

This is a multi part 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 a coin tossed three times is $S = (HHH, HHT, HTH, HTT, THH, THT, TTH, TTT)$
Total number of outcomes in S = 8.
For our event E: atleast one heads occur, $\rightarrow E = (HHH, HHT, HTH, THH, HTT, THT, TTH)$. Total number of outcomes in E = 7 .
For our event F: atleast 1 tail occurs $\rightarrow F = (TTT, THT, TTH, HTT, HHT, HTH, THH). The total number of outcomes in F = 7.$\Rightarrow P(E) = \large \frac{\text{Number of favorable outcomes in E}}{\text{Total number of outcomes in S}} = \frac{7}{8}\Rightarrow P(F) = \large \frac{\text{Number of favorable outcomes in F}}{\text{Total number of outcomes in S}} = \frac{7}{8}$For our event (E \;\cap\; F) = (HHT, HTH, THH, HTT, THT, TTH)$
Total number of favorable outcome is 6.
$\Rightarrow P(E\;\cap\;F) = \large \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes in S}} = \frac{6}{8}$
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{6}{8}} {\Large \frac{7}{8}}$ = $\large \frac{6}{7}$