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(E|F).
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: Coin is tossed three times, therefore total outcomes are S= {HHH, HHT, THH, HTH, TTH, THT, HTT, TTT}. Use the formula, $P(E)={\displaystyle \frac{PossibleOutcomes}{TotalOutcomes}}$ and $P(E|F)={\displaystyle \frac{P(E\bigcap F)}{P(F)}}$ to find the solution.

Complete step-by-step answer:
Coins are tossed three times.
S= {HHH, HHT, THH, HTH, TTH, THT, HTT, TTT}
(i) E: head on third toss
E= {HHH, HTH, THH, TTH}
$P(E)={\displaystyle \frac{PossibleOutcomes}{TotalOutcomes}}$
$\therefore P(E)={\displaystyle \frac{4}{8}}={\displaystyle \frac{1}{2}}$
F: heads on first two tosses
F= {HHH, HHT}
$\therefore P(F)={\displaystyle \frac{2}{8}}={\displaystyle \frac{1}{4}}$
Therefore, $P(E|F)={\displaystyle \frac{P(E\bigcap F)}{P(F)}}$
Now, $$\begin{gathered} E\bigcap F=\{HHH\} \\ P(E\bigcap F)={\displaystyle \frac{1}{8}} \end{gathered}$$
Using the equation, $$\begin{gathered} P(E|F)={\displaystyle \frac{P(E\bigcap F)}{P(F)}}={\displaystyle \frac{{\displaystyle \frac{1}{8}}}{{\displaystyle \frac{1}{4}}}}={\displaystyle \frac{1}{2}} \\ P(E|F)=0.50 \end{gathered}$$
(ii) E: at least two heads
Coins are tossed three times.
S= {HHH, HHT, THH, HTH, TTH, THT, HTT, TTT}
E= {HHT, THH, HTH, HHH}
$P(E)={\displaystyle \frac{4}{8}}={\displaystyle \frac{1}{2}}$
F: at most two heads
F = {HHT, THH, HTH, TTH, THT, HTT, TTT}
$P(F)={\displaystyle \frac{7}{8}}$
Also, $$\begin{gathered} E\bigcap F=\{HHT,THH,HTH\} \\ P(E\bigcap F)={\displaystyle \frac{3}{8}} \end{gathered}$$
Now, $$\begin{gathered} P(E|F)={\displaystyle \frac{P(E\bigcap F)}{P(F)}}={\displaystyle \frac{{\displaystyle \frac{3}{8}}}{{\displaystyle \frac{7}{8}}}}={\displaystyle \frac{3}{7}} \\ P(E|F)=0.42 \end{gathered}$$
(iii) E: at most two tails
Coins are tossed three times.
S= {HHH, HHT, THH, HTH, TTH, THT, HTT, TTT}
E= {HHH, HHT, HTH, THH, TTH, THT, HHT}
$P(E)={\displaystyle \frac{7}{8}}$
F: at least one tail
F = {HHT, HTH, THH, TTH, THT, HTT, TTT}
$P(F)={\displaystyle \frac{7}{8}}$
Also, $$\begin{gathered} E\bigcap F=\{HHT,THH,HTH,TTH,THT,HTT\} \\ P(E\bigcap F)={\displaystyle \frac{6}{8}} \end{gathered}$$

Now,
$$\begin{gathered} P(E|F)={\displaystyle \frac{P(E\bigcap F)}{P(F)}}={\displaystyle \frac{{\displaystyle \frac{6}{8}}}{{\displaystyle \frac{7}{8}}}}={\displaystyle \frac{6}{7}} \\ P(E|F)=0.85 \end{gathered}$$
So, the correct option is Option (B).

Note: Whenever such a type of question appears note down all the outcomes of the event and the possible outcomes in the particular given case, as given in question is the coin is tossed 3 times. And then find the probability of all the cases using the standard formula.