Respuesta :
Probability of an event measures its chance of occurrence. P(Green in first draw) = 5/14, P(Red in second draw) = 2/14. P(Green in first and red in second draw) = 5/98
How to calculate the probability of an event?
Suppose that there are finite elementary events in the sample space of the considered experiment, and all are equally likely.
Then, suppose we want to find the probability of an event E.
Then, its probability is given as
[tex]P(E) = \dfrac{\text{Number of favorable cases}}{\text{Number of total cases}} = \dfrac{n(E)}{n(S)}[/tex]
where favorable cases are those elementary events who belong to E, and total cases are the size of the sample space.
What is chain rule in probability?
For two events A and B, by chain rule, we have:
[tex]P(A \cap B) = P(B)P(A|B) = P(A)P(B|A)[/tex]
where P(A|B) is probability of occurrence of A given that B already occurred.
How to find if two events are independent?
Suppose that two events are denoted by A and B.
They are said to be independent event if and only if:
[tex]P(A \cap B) = P(A)P(B)[/tex]
It is because
[tex]P(A|B) = P(A)\\P(B|A) = P(B)[/tex]
and therefore, using chain rule and above facts gives:
[tex]P(A \cap B) = P(A)P(B)[/tex]
The question is missing some informations, which are:
- Total marbles in the bag = 14
- Total red marbles = 5
- Total green marbles = 2
- rest of the 7 marbles are of some other colors.
The result of first draw and the second draw are independent of each other as Deepak replaces the marble after first draw, therefore, making no changes in the count of balls of each color in the considered bag, which decides the probability needed.
Now, let we take:
A = probability that Deepak picks green ball in first draw
B = probability that Deepak picks red ball in the second draw
Then, we get:
[tex]P(A) =\dfrac{5}{14}\\\\P(B) = \dfrac{2}{14}\\\\P(A \cap B) = P(A)P(B) = \dfrac{5 \times 2}{14^2} = \dfrac{5}{98}[/tex]
(there are 14 ways to select a ball, but 5 favorable ball for event A to occur, and 2 favorable balls for event B to occur,therefore the aforesaid probabilities.)
Thus, for this case, we get:
- P(Green in first draw) = 5/14,
- P(Red in second draw) = 2/14.
- P(Green in first and red in second draw) = 5/98
Learn more about probability here:
brainly.com/question/1210781
