 QUESTION

# Two coins are tossed once, where(i) E: tail appears on one coin, F: one coin shows head(ii) E: no tail appears, F: no head appearsDetermine $P\left( {E|F} \right)$A.1,0B.1,0.2C.0,2D.1,3

Hint- In this particular type of question we need to proceed by finding the sample space (S) of the event. Then we need to find P(E) and P(F) in both cases and then use the formula of conditional probability to find $P\left( {E|F} \right)$.

S (sample space) =$\left\{ {\left( {h,h} \right),\left( {h,t} \right),\left( {t,h} \right)\left( {t,t} \right)} \right\}$ = 4
$E = \left\{ {\left( {h,t} \right),\left( {t,h} \right)} \right\} \\ P\left( E \right) = \dfrac{2}{4} = \dfrac{1}{2} \\ F = \left\{ {\left( {h,t} \right),\left( {t,h} \right)} \right\} \\ P\left( F \right) = \dfrac{2}{4} = \dfrac{1}{2} \\ Also,E \cap F = \left\{ {\left( {h,t} \right),\left( {t,h} \right)} \right\} \\ P\left( {E \cap F} \right) = \dfrac{2}{4} = \dfrac{1}{2} \\ P\left( {E|F} \right) = \dfrac{{P\left( {E \cap F} \right)}}{{P\left( F \right)}} = \dfrac{{\dfrac{1}{2}}}{{\dfrac{1}{2}}} = 1 \\$
$E = \left\{ {\left( {h,h} \right)} \right\} \\ P\left( E \right) = \dfrac{1}{4} \\ F = \left\{ {\left( {t,t} \right)} \right\} \\ P\left( F \right) = \dfrac{1}{4} \\ Also,E \cap F = \phi \\ so{\text{ P}}\left( {E \cap F} \right) = 0 \\ now{\text{ P}}\left( {E|F} \right) = \dfrac{{{\text{P}}\left( {E \cap F} \right)}}{{P\left( F \right)}} = \dfrac{0}{{\dfrac{1}{4}}} = 0 \\$
We have to recall that conditional probability is a measure of the probability of an event occurring given that another event has occurred. Example, if a person has dengue, they might have a 90% chance of testing positive for dengue. In this case what is being measured is that if event B (having dengue) has occurred, the probability of A (test is positive) given that B (having dengue) occurred is 90% i.e. $P\left( {A|B} \right)$ = 90%.