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

Answer 1

Since the coin is tossed three times. Then, find the total outcomes of the event.
In part (i), the possible outcomes for E are the head on the third toss, no matter what is on the first 2 tosses, and for F is head on the first 2 tosses neglecting the last toss result.
In part (ii), the possible outcomes for E are either two heads or three heads and for F is no 3 heads occur.
In part (iii), the possible outcomes for E are no 3 tails and for F is no 3 heads occur.

Complete step-by-step solution:
It is given that a coin is tossed three times.
\Rightarrow S = \left\\{ {HHH,HHT,HTH,THH,HTT,THT,TTH,TTT} \right\\}
So, the total outcomes will be,
$\Rightarrow n\left( S \right) = 8$
(i)
For E, the favorable outcome is,
\Rightarrow E = \left\\{ {HHH,HTH,THH,TTH} \right\\}
The number of favorable outcomes is,
$\Rightarrow n\left( E \right) = 4$
So, the probability is,
$P\left( E \right) = \dfrac{{n\left( E \right)}}{{n\left( S \right)}}$
Substitute the values,
$\Rightarrow P\left( E \right) = \dfrac{4}{8}$
Cancel out the common factor,
$\therefore P\left( E \right) = \dfrac{1}{2}$
For F, the favorable outcome is,
\Rightarrow F = \left\\{ {HHH,HHT} \right\\}
The number of favorable outcomes is,
$\Rightarrow n\left( F \right) = 2$
So, the probability is,
$P\left( F \right) = \dfrac{{n\left( F \right)}}{{n\left( S \right)}}$
Substitute the values,
$\Rightarrow P\left( F \right) = \dfrac{2}{8}$
Cancel out the common factor,
$\therefore P\left( F \right) = \dfrac{1}{4}$
Hence, the probability of E is $\dfrac{1}{2}$ and of F is $\dfrac{1}{4}$.
(ii)
For E, the favorable outcome is,
\Rightarrow E = \left\\{ {HHH,HHT,HTH,THH} \right\\}
The number of favorable outcomes is,
$\Rightarrow n\left( E \right) = 4$
So, the probability is,
$P\left( E \right) = \dfrac{{n\left( E \right)}}{{n\left( S \right)}}$
Substitute the values,
$\Rightarrow P\left( E \right) = \dfrac{4}{8}$
Cancel out the common factor,
$\therefore P\left( E \right) = \dfrac{1}{2}$
For F, the favorable outcome is,
\Rightarrow F = \left\\{ {HHT,HTH,THH,HTT,THT,TTH,TTT} \right\\}
The number of favorable outcomes is,
$\Rightarrow n\left( F \right) = 7$
So, the probability is,
$P\left( F \right) = \dfrac{{n\left( F \right)}}{{n\left( S \right)}}$
Substitute the values,
$\therefore P\left( F \right) = \dfrac{7}{8}$
Hence, the probability of E is $\dfrac{1}{2}$ and of F is $\dfrac{7}{8}$.
(iii)
For E, the favorable outcome is,
\Rightarrow E = \left\\{ {HHH,HHT,HTH,THH,TTH,THT,HTT} \right\\}
The number of favorable outcomes is,
$\Rightarrow n\left( E \right) = 7$
So, the probability is,
$P\left( E \right) = \dfrac{{n\left( E \right)}}{{n\left( S \right)}}$
Substitute the values,
$\therefore P\left( E \right) = \dfrac{7}{8}$
For F, the favorable outcome is,
\Rightarrow F = \left\\{ {HHT,HTH,THH,HTT,THT,TTH,TTT} \right\\}
The number of favorable outcomes is,
$\Rightarrow n\left( F \right) = 7$
So, the probability is,
$P\left( F \right) = \dfrac{{n\left( F \right)}}{{n\left( S \right)}}$
Substitute the values,
$\therefore P\left( F \right) = \dfrac{7}{8}$
Hence, the probability of E and F is $\dfrac{7}{8}$.

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 the question the coin is tossed 3 times. And then find the probability of all the cases using the standard formula.