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, $P(E) = \dfrac{{Possible Outcomes}}{{Total Outcomes}}$ and $P(E|F) = \dfrac{{P(E\bigcap {F)} }}{{P(F)}}$ 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}
$P(E) = \dfrac{{Possible Outcomes}}{{Total Outcomes}}$
$\therefore P(E) = \dfrac{4}{8} = \dfrac{1}{2}$
F: heads on first two tosses
F= {HHH, HHT}
$\therefore P(F) = \dfrac{2}{8} = \dfrac{1}{4}$
Therefore, $P(E|F) = \dfrac{{P(E\bigcap {F)} }}{{P(F)}}$
Now, $
  E\bigcap F = \{ HHH\} \\
  P(E\bigcap {F) = \dfrac{1}{8}} \\
 $
Using the equation, $
  P(E|F) = \dfrac{{P(E\bigcap {F)} }}{{P(F)}} = \dfrac{{\dfrac{1}{8}}}{{\dfrac{1}{4}}} = \dfrac{1}{2} \\
  P(E|F) = 0.50 \\
    \\
 $
(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}
$P(E) = \dfrac{4}{8} = \dfrac{1}{2}$
F: at most two heads
F = {HHT, THH, HTH, TTH, THT, HTT, TTT}
$P(F) = \dfrac{7}{8}$
Also, $
  E\bigcap F = \{ HHT,THH,HTH\} \\
  P(E\bigcap {F) = \dfrac{3}{8}} \\
$
Now, $
  P(E|F) = \dfrac{{P(E\bigcap {F)} }}{{P(F)}} = \dfrac{{\dfrac{3}{8}}}{{\dfrac{7}{8}}} = \dfrac{3}{7} \\
  P(E|F) = 0.42 \\
    \\
 $
(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}
$P(E) = \dfrac{7}{8}$
F: at least one tail
F = {HHT, HTH, THH, TTH, THT, HTT, TTT}
$P(F) = \dfrac{7}{8}$
Also, $
  E\bigcap F = \{ HHT,THH,HTH,TTH,THT,HTT\} \\
  P(E\bigcap {F) = \dfrac{6}{8}} \\
 $

Now,
$
  P(E|F) = \dfrac{{P(E\bigcap {F)} }}{{P(F)}} = \dfrac{{\dfrac{6}{8}}}{{\dfrac{7}{8}}} = \dfrac{6}{7} \\
  P(E|F) = 0.85 \\
    \\
 $
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.