Question

A game consists of tossing a coin 3 times and noting its outcome. A boy wins if all tosses give the same outcome and loses otherwise. Find the probability that the boy loses the game.

Answer 1

Hint: In this problem first we need to find the possible number of outcomes in tossing a coin three times and then calculate the number of ways in which a boy loses the game i.e. condition when all the outcomes are not the same.

Complete step-by-step answer:
The possible outcomes on tossing a coin 3 times are=
$
   \Rightarrow {\text{ \{ (H,H,H),}} \\
  {\text{ (H,H,T),}} \\
  {\text{ (H,T,H),}} \\
  {\text{ (T,H,H),}} \\
  {\text{ (T,T,H),}} \\
  {\text{ (T,H,T),}} \\
  {\text{ (H,T,T),}} \\
  {\text{ (T,T,T)\} }} \\
    \\
    \\
    \\
 $
Thus, number of possible outcomes are = $8$
It is given that a boy wins if all tosses give the same outcome i.e.
                                                                                            =$
  \{ ({\text{H,H,H)}} \\
  {\text{ }}({\text{T,T,T)\} }} \\
 $
Thus number of ways in which a boy wins =$2$
Then, number of ways in which a boy loses are =Total- number of ways in which a boy wins
                                                                                     $
   = 8 - 2 \\
   = 6 \\
$
Let ${\text{P(E)}}$ be the probability that loses the game
${\text{P(E) = }}$$\dfrac{{{\text{Number of ways in which a boy loses game}}}}{{{\text{Total number of possible outcomes}}}}$
$
   \Rightarrow {\text{P(E) = }}\dfrac{6}{8} \\
   \Rightarrow {\text{P(E) = }}\dfrac{3}{4} \\
$
Hence, the probability that a boy loses the game is $\dfrac{3}{4}$.

Note: Whenever you find this type of problem the key concept of solving the problem that firstly determine the total number of outcomes then determine the number of favourable outcomes then use the formula of finding probability ${\text{P(E) = }}\dfrac{{{\text{Number of favorable outcomes}}}}{{{\text{Number of total outcomes}}}}$.