Probability is a calculation of uncertainty of actions in a random experiment. The two events are called as Independent if and only if the outcomes of initial event does not alter the outcomes of next event.
Let R and S are the two events, the occurrence of the R event which does not alter the occurrence of S event, and then it is referred as Independent.
According to probability formula P(R and S) = P(R) • P(S)
Problems Based on Probability Independent:
Pro 1: In a car parking, there are 5 yellow cars, 4 blue cars and 2 black cars are parked. If a car is chosen at random, What is the probability of choosing yellow and black car ?
Sol : Probability of choosing yellow car P(Y) = 5/11
Probability of choosing black car P(B) = 2/11
So, P( Y and B ) = 5/11 • 2/11
Pro 2: A Bag contains 5 red, 3 green, 3 blue and 8 yellow chocolates. A Chocolate is chosen at random from the bag. After replacing it, a second chocolate is chosen. What is the probability of choosing a green and a yellow chocolate?
Sol : Probability of taking green P(G) = 3/19
Probability of taking yellow P(Y) = 8/19
So, P(G and Y) = 3/19 • 8/19
= 24 / 361.
Example for Events that Occur in Sequence:
Ex 1: In a School survey they found that 85% of students like Math subject. If 3 students are selected at random, what is the probabilit that all three like Math?
Sol : Probability of student 1 likes Math P(1) = 0.85
Probability of student 2 likes Math P(2) = 0.85
Probability of student 3 likes Math P(3) = 0.85
So, probability that all three like Math = (0.85) • (0.85) • (0.85)
In percent = 61%
Ex 2: Consider 8 out of 9 people having Dog in their Home. If 4 people are selected at random with replacement, what is the probability that all four having Dogs?
Sol : Probability of person 1 having Dog P(1) = 8/9
Probability of person 2 having Dog P(2) = 8/9
Probability of person 3 having Dog P(3) = 8/9
Probability of person 4 having Dog P(4) = 8/9
So, Probability of all four having Dogs = P(1) • P(2) • P(3) • P(4)
= 8/9 • 8/9 • 8/9 • 8/9
= 4096 / 6561