Question

A bag containing 4 red and 4 blue balls. Four balls are drawn one by one from the bag, then find the probability that the drawn balls are in alternate color.

Answer 1

Probability means possibility. It is a branch of mathematics that deals with the occurrence of events. 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.
Event is a subset of the sample space i.e. a set of outcomes of the random experiment.
${\text{Probability of an event}} = \dfrac{{{\text{Number of occurrence of event A in S}}}}{{{\text{Total number of cases in S}}}}{\text{ }}$
$= \dfrac{{n\left( A \right)}}{{n\left( S \right)}}$
If the probability of occurrence of an event A is not affected by the occurrence of another event B, then A and B are said to be independent events.
If A and B are two independent events for a random experiment, then the probability of simultaneous occurrence of two independent events will be equal to the product of their probabilities. Hence,
$P(A \cap B) = P(A).P(B)$
Students should use the multiplication theorem for independent events. Here the four balls are drawn one by one so, second drawn is independent of 1st drawn.

Complete step by step solution:
Given,
Number of red balls = 4
Number of blue balls = 4
Here the ball is drawn at random one by one
For the ball drawing in alternate color there are 2 cases.
Case 1: Red, Blue, Red, Blue
Case 2: Blue, Red, Blue, Red
Since drawing of 2nd ball is independent of 1st ball
Hence they are independent events
Hence by multiplication theorem
$P(E) = P(RBRB) + P(BRBR)$
$= \left( {\dfrac{4}{8}} \right)\left( {\dfrac{4}{8}} \right)\left( {\dfrac{4}{8}} \right)\left( {\dfrac{4}{8}} \right) + \left( {\dfrac{4}{8}} \right)\left( {\dfrac{4}{8}} \right)\left( {\dfrac{4}{8}} \right)\left( {\dfrac{4}{8}} \right) \\\ = \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2} + \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2} \\\ = \dfrac{1}{{16}} + \dfrac{1}{{16}} \\\$
$P(E) = \dfrac{1}{8}$
Hence, Probability of getting alternate color ball $= \dfrac{1}{8}$

Note:
Here in this question, there is use of the concept of independent events. Since the ball is drawn one by one then we have to use a separate formula one main thing is that there are two cases of alternate colors. There are some main rules associated with basic probability:
1. P(not A)=1-P(A)
2. P(A or B)=P(event A occurs or event B occurs or both occur)
3. P(A and B)=P(both event A and event B occurs)
4. The general Addition Rule: P(A or B) = P(A) + P(B) – P(A and B)