Question

A coin is tossed three times, where
 (i) $E$:head on third toss, $F$: heads on first two tosses
(ii) $E$:at least two heads, $F$: at most two heads
(iii) $E$:at most two tails, $F$: at least one tail Determine $P\left( {E|F} \right)$
(A) $0.42,0.50,0.85$
(B) $0.50,0.42,0.85$
(C) $0.85,0.42,0.30$
(D) $0.42,0.46,0.47$

Answer 1

Hint: In the given question we are given that a coin is tossed three tunnes. Then two events are recorded recorded $E$ and $F$ and there events $E$ and $F$ have some different outcomes then we have to find out $P(E/F)$ for every case where $P(E/F)$ is conditional probability for $2$ events $E$ and $F$ where $F$ is given event $E$ and is event is to be found.

Complete step-by-step answer:
In the questions, coin is tossed three times and let $E$ and $F$ be two events that records different for there $2$ events and for there three recorded different two events $E$ and $F$ we have to find out the value of $P(E/F)$) for each case where $P(E/F)$ is conditional probability of $E$ to $F$ where $E$, is the unknown event and $F$ is the known or given event.
The identity for conditional probability is
$P(E/F) = \dfrac{{p(E \cap F)}}{{p(F)}}$
Which \[p(E \cap F)\] is intersection of $E$ and $F$ means the Common event from $E$ and $F$
Also a coin is tossed \[3\] times so the sample space
Will became $\left\{ {\left. {HHH,HHT,HTH,THH,HTT,TTH,THT,TTT} \right\}} \right.$
It include $8$ elements Nbw up will solve three parts one by one
(i) $E = \left\{ {\left. {HHH,HTH,THH,TTH} \right\}} \right.$
because head on third toss, rest 2 be of any
$E = \left\{ {\left. {HHH,HHT} \right\}} \right.$
Because head on first two
$\therefore P(f) = \dfrac{2}{8} = \dfrac{1}{4}$
and $P(E \cap F) = \dfrac{1}{8}$
because only 1 is common in $E$ and $F$
$\therefore P(E/F) \Rightarrow \dfrac{{\dfrac{1}{8}}}{{\dfrac{1}{4}}} = \dfrac{4}{8} = \dfrac{1}{2}$
Which gives \[0.50\] (ii) $E = \left\{ {\left. {HHH,HHT,HTH,THH} \right\}} \right.$
because at least two heads, three may be there
\[F = \left\{ {\left. {HHT,HTH,HTT,THH,THT,TTH,TTT} \right\}} \right.\]
because at most two heads, zero or one be there
$\therefore P(E \cap F) \Rightarrow \dfrac{3}{8}{\text{ and }}P(F) = \dfrac{7}{8}$
on substitute in formula use get
$\therefore P(E/F) \Rightarrow \dfrac{{P(E \cap F)}}{{P(F)}} = \dfrac{{\dfrac{2}{8}}}{{\dfrac{7}{8}}}$
 $ = \dfrac{3}{7} = 0.42$
(iii) $E = \left\{ {\left. {HHH,HHT,HTT,HTH,THH,THT,TTH} \right\}} \right.$
because at least two tails, so zero or one may be there
$F = \left\{ {\left. {HHT,HTT,HTH,THH,THT,TTH,TTT} \right\}} \right.$
because at least two tails, so two or three may be there
$\therefore P(F) = \dfrac{7}{8}{\text{ and }}P(E \cap F) = \dfrac{6}{8}$
On substituting in formulae up get
$P(E/F) = \dfrac{{P(E \cap F)}}{{P(F)}} = \dfrac{{\dfrac{6}{8}}}{{\dfrac{7}{8}}}$
$ \Rightarrow \dfrac{6}{7} = 0.85$
So, option (B) is correct.

Note: Probability of any event lies within the range of $0$ to $1$ If means probability of any null event is zero and probability of whole sample space is one also probability of any event can not be negative but probability of any event can be fractional which includes the limit of probability from $0$ to $1$.