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

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- 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}}$ = $\large \frac{6}{7}$

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