Probability Independent

Probability is a calculation of uncertainty of actions in a random experiment. The two events are said to be probability Independent if and only if the occurrence of initial event does not alter the occurrence 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  probability Independent.

P(R and S) = P(R) . P(S)

I like to share this Probability Tree Diagram with you all through my article.

Problems on Probability Independent:

Problems on Probability Independent are as follows:

1) In a car parking, there are 2red cars, 4blue cars and 6 black cars are parked. A car is chosen at random, what is the probability of choosing red and black car in second turn?

Solution:

Probability of choosing red car P(R)       = 2/12 =1/6

Probability of choosing black car P(B)    = 6/12 =1/2

So, P(R and B)    = 1/6 * 1/2

= 1/12.

2) A Bag contains 6 red, 5 green, 2 blue and 4 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?

Solution:

Probability of taking green P(G)           = 5/17

Probability of taking yellow P(Y)           = 4/17

So, P(G and Y)            = 5/17 * 4/17

= 20/289.

Events occur in sequence on probability independent:

1) In a School survey they found that 79% of students like Math subject. If 3 student are selected at random, what is the probability that all three like Math?

Solution:

Probability of student 1 likes Math P(1)  =  0.79

Probability of student 2 likes Math P(2)  =  0.79

Probability of student 3 likes Math P(3)  =  0.79

So, probability that all three like Math  = (0.79) * (0.79) * (0.79)

= 0.49

As percent   =49%

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2) Consider 7 out of 10 people having Dog in their Home. If 4 people are selected at random with replacement, what is the probability that all four having Dogs?

Solution:

Probability of people 1 having Dog P(1) = 7/10

Probability of people 2 having Dog P(2) = 7/10

Probability of people 3 having Dog P(3) = 7/10

Probability of people 4 having Dog P(4) = 7/10

So, Probability of all four having Dogs = P(1) * P(2) * P(3) * P(4)

= 7/10 * 7/10 * 7/10 * 7/10

= 2401/10000