Respuesta :
Answer:
First of all, we do not know if the box is open or closed. If the box is open and the person drawing the ball can see them, the answer is 50/50 since there are only two colors to choose from. If the box was closed and the person drawing the ball could not see the colors of the balls, the odds of drawing a red ball would mathematically be 1 in 3 because there were 6 balls in all with 2 of them red (6:2 = 3:1). The second ball drawn -- the one from box B -- has no bearing on these odds. As usual, I could be wrong about this and would appreciate knowing the correct answer or, if I am in fact correct, I would also like to know that.
Step-by-step explanation:
The probability that the first ball drawn was red given that the second ball drawn was red is 0.1167 approx
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}}[/tex]
where favorable cases are those elementary events who belong to E, and total cases are the size of the sample space.
What is law of total probability?
Suppose that the sample space is divided in n mutual exclusive and exhaustive events tagged as
[tex]B_i \: ; i \in \{1,2,3.., n\}[/tex]
Then, suppose there is event A in sample space.
Then probability of A's occurrence can be given as
[tex]P(A) = \sum_{i=1}^n P(A \cap B_i)[/tex]
Using the chain rule, we get
[tex]P(A) = \sum_{i=1}^n P(A \cap B_i) = \sum_{i=1}^n P(A)P(B_i|A) = \sum_{i=1}^nP(B_i)P(A|B_i)[/tex]
What is the chain rule in probability for two events?
For two events A and B:
The chain rule states that the probability that A and B both occur is given by:
[tex]P(A \cap B) = P(A)P(B|A) = P(B)P( A|B)[/tex]
Let we take the updated box B as B' when we add a ball from box A to box B
For the given condition, let we consider two events as:
E = Event of drawing a red ball from box A
F = Event of drawing a red ball from box B'
Now B' can be one of the two states. One state being that the first ball putted in box B is blue ball, and other state being the first ball putted in box B is red ball.
P(A red ball was added to B) = P(E)
P(A blue ball was added to B) = 1 - P(E) (complement event)
The probability that the first ball drawn was red given that the second ball drawn was red is the probability of occurrence of E when F is given to be occurred.
This is symbolically written as [tex]P(E|F)[/tex]
We get P(E) = 2/6 (since favorable event is drawing red ball, which can be done in 2 ways as there are 2 reds in box A, and total ways of drawing a ball is 6).
From the chain rule of probability, we get:
[tex]P(E)P(F|E)= P(F)P(E|F)\\or\\P(E|F) = \dfrac{P(E)P(F|E)}{P(F)}[/tex]
P(F|E) = probability of F's happening when E has occurred.
Since it is given that E has occurred for above case, that means there will be 7 red balls in second box. And thus,
P(F|E) = 7/10 (total ways of drawing a ball from modified box having extra red= 10, total ways of drawing a red from second box modified having extra red= 7)
Also, from law of total probability and chain rule, we get:
P(F) = P(E)P(F|E) + P(E')P(F|E')
where E' = complement event of E, that means event of drawing blue ball from first box.
Thus, P(E') = 4/6 (from same logic of finding the probability)
and P(F|E') = 4/10( since now in modified second box, there are 4 blues and total 10 balls)
Thus, we get:
P(F) = P(E)P(F|E) + P(E')P(F|E') = 2/6 × 7/10 + 4/6 × 4/10 = 0.5
Using this value, and the equation [tex]P(E|F) = \dfrac{P(E)P(F|E)}{P(F)}[/tex] and P(E)'s vlue and P(F|E)'s value, we get:
[tex]P(E|F) = \dfrac{P(E)P(F|E)}{P(F)} = \dfrac{2/6 \times 7/10}{0.5} \approx 0.1167[/tex]
Thus, the probability that the first ball drawn was red given that the second ball drawn was red is 0.1167 approx
Learn more about probability here:
brainly.com/question/1210781
