Question

A coin is tossed five times, find the probability of getting 4 tails.
A. $\dfrac{5}{{32}}$
B. $\dfrac{1}{{32}}$
C. $0.4$
D. $0.25$ .

Answer 1

Hint:
Probability means possibility. It is a branch of mathematics that deals with the occurrence of event. It is used to predict how likely events are to happen. To find the probability of a single event to occur, first, we should know the total number of possible outcomes.
Probability can range in from 0 to 1, where 0 means the event to be an impossible one and 1 indicates a certain event.
Random experiment is an experiment where we know the set of all possible outcomes but find it impossible to predict one at any particular execution.
Sample space is defined as the set of all possible outcomes of a random experiment. Example: Tossing a head, Sample Space(S) = {H, T}. The probability of all the events in a sample space adds up to 1.
Event is a subset of the sample space i.e. a set of outcomes of the random experiment.
\[{\text{Probability of an event}} = \dfrac{{{\text{Number of occurrence of event A in S}}}}{{{\text{Total number of cases in S}}}}{\text{ }}\]
\[ = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]

 Complete step by step solution:
The number of Sample spaces when a coin is tossed is 2, head and tail.
Here a coin is tossed 5 times
∴ The sample space of tossing 5 times
$
n(S) = 2 \times 2 \times 2 \times 2 \times 2 \\
n(S) = 32 \\
$
Now, the favorable cases of getting 4 tails
Let H denotes head and T denotes tail
\[
E{\text{ }} = {\text{ }}\left\{ {HTTTT,{\text{ }}THTTT,{\text{ }}TTHTT,{\text{ }}TTTHT,{\text{ }}TTTTH} \right\} \\
n(E) = 5 \\
\]
We have the probability of occurrence of an event
$
P(E) = \dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of outcomes}}}} \\
P(E) = \dfrac{{n(E)}}{{n(S)}} \\
P(E) = \dfrac{5}{{32}} \\
$
Hence, the required probability of getting 4 tails when a coin is tossed 5 times is $\dfrac{5}{{32}}$
∴Option (A) is correct.

 Note:
There are some main rules associated with basic probability:
1. For any event A, 0≤P(A)≤1.
2. The sum of the probabilities of all possible outcomes is 1.
3. P(not A)=1-P(A)
4. P(A or B)=P(event A occurs or event B occurs or both occur)
5. P(A and B)=P(both event A and event B occurs)