This is part of a multi-part question in the textbook that is answered separately here.

- 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: a head occurs on the third toss, $\rightarrow E = (HHH, THH, HTH, TTH)$. Total number of outcomes in E = 4 .

For our event F: heads on the first two tosses $\rightarrow F = (HHH, HHT). The total number of outcomes in F = 2.

$\Rightarrow P(E) = \large \frac{\text{Number of favorable outcomes in E}}{\text{Total number of outcomes in S}} = \frac{4}{8} = \frac{1}{2}$

$\Rightarrow P(F) = \large \frac{\text{Number of favorable outcomes in F}}{\text{Total number of outcomes in S}} = \frac{2}{8} = \frac{1}{4}$

For our event E: a head occurs on the third toss, $\rightarrow E = (HHH, THH, HTH, TTH)$ and F: heads on the first two tosses $\rightarrow F = (HHH, HHT) \rightarrow (E \;\cap\; F) = (HHH)$

For our event (E \;\cap\; F) = (HHH)$. Total number of favorable outcome is 1.

$\Rightarrow P(E\;\cap\;F) = \large \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes in S}} = \frac{1}{8}$

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{1}{8}} {\Large \frac{1}{4}}$ = $\large \frac{1}{2}$

Ask Question

Tag:MathPhyChemBioOther

Take Test

...