# Solve Probability Independent

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

= 10/121.

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)

= 0.61

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