Simple Probability Problems

Introduction to simple probability problems:

Simple probability is the numerical measure of the likelihood of an event to occur. If in an observation there are n possible ways exhaustive and mutually exclusive and out of them in m ways in the event A occurs, then the probability of occurrence of the event A is given by P(A) = m / n.

If in a random sequence of n trials of event, M are favorable to the event, then the probability of that event occurring is the limit of the ratio M / n.   When n is very large, this lies between 0 and 1.P (A) = 0, means that the event cannot take place. P (A) = 1 means the event is bound to occur. 

Simple Probability Problems - Experiment:
The simple probability is the study of chance or likelihood of an event to happening. The directly or indirectly, simple probability plays a role in the all activities.

Experiment 1 : Tossing a coin

Possible outcomes are head or tail.

Sample space, S = {head, tail}.

Experiment 2: Tossing a die

Possible outcomes are the numbers 1, 2, 3, 4, 5, and 6

Sample space, S = {1, 2, 3, 4, 5, 6}

Simple Solved Probability Problems:

Problem 1:

Find the probability of getting a head when a coin is tossed once. Also, find the probability of getting a tail.

Solution:

In the experiment of tossing a coin once, the number of possible outcomes is two — Head (H) and Tail (T). Let E be the event ‘getting a head’. The number of outcomes favorable to E, (i.e., of getting a head) is 1. Therefore,

P (E) = P (head) = (Number of outcomes favorable to E) / (Number of all possible outcomes)

P (E)   = 1/ 2

Similarly, if F is the event ‘getting a tail’, then

P (F) = P (tail) = 1 / 2.

Problem 2:

Suppose we throw a die once.

(i)                What is the probability of getting a Number greater than 4?

(ii)               What is the probability of getting a number less than or equal to 4 ?

Solution:

(i) Here, let E be the event ‘getting a number greater than 4’. The number of possible outcomes is six: 1, 2, 3, 4, 5 and 6, and the outcomes favorable to E are 5 and 6. Therefore, the number of outcomes favorable to E is 2. So,

P (E) = P (number greater than 4) = 2 / 6 = 1 / 3

(ii) Let F be the event ‘getting a number less than or equal to 4’.

Number of possible outcomes = 6

Outcomes favorable to the event F are 1, 2, 3, 4

So, the number of outcomes favorable to F is 4.

Therefore, P (F) = 4 / 6 = 2 / 3          

The event E has 2 outcomes and the event F has 4 outcomes.