This exercise is about conditional probability.

A jar contains five red balls numbered 1 to 5, and six green balls numbered 1 to 6. A ball is drawn at random from the jar. Find the following conditional probabilities. (Enter your probabilities as fractions.)
(a) The ball is red, given that it is numbered 2.


(b) The ball is green, given that it is numbered 6.


(c) The ball is red, given that it has an even number.


(d) The ball has an even number, given that it is red

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To solve for the conditional probabilities, we need to use the definition of conditional probability:

P(A|B) = P(A ∩ B) / P(B)

Where P(A|B) denotes the probability of event A occurring given that event B has occurred, P(A ∩ B) denotes the probability of both events A and B occurring, and P(B) denotes the probability of event B occurring.

Now, let's solve for each of the given conditional probabilities step by step:

(a) The ball is red, given that it is numbered 2.

To find this conditional probability, we first need to find the probability of both events occurring: the ball being red and numbered 2. From the information provided, there is only one red ball numbered 2, and there are a total of 11 balls in the jar.

P(Red and numbered 2) = 1/11

Next, we need to find the probability of the event B occurring, which is the ball being numbered 2. From the information provided, there are a total of 11 balls in the jar.

P(Numbered 2) = 1/11

Finally, we can calculate the conditional probability using the formula:

P(Red|Numbered 2) = P(Red and numbered 2) / P(Numbered 2)

P(Red|Numbered 2) = (1/11) / (1/11)

P(Red|Numbered 2) = 1/1 = 1

So, the conditional probability that the ball is red, given that it is numbered 2, is 1.

(b) The ball is green, given that it is numbered 6.

Using the same approach as above, we start by finding the probability of both events occurring: the ball being green and numbered 6. There is only one green ball numbered 6, and there are a total of 11 balls in the jar.

P(Green and numbered 6) = 1/11

Next, we find the probability of the event B occurring, which is the ball being numbered 6. From the information provided, there are a total of 11 balls in the jar.

P(Numbered 6) = 1/11

Then, we can calculate the conditional probability:

P(Green|Numbered 6) = P(Green and numbered 6) / P(Numbered 6)

P(Green|Numbered 6) = (1/11) / (1/11)

P(Green|Numbered 6) = 1/1 = 1

So, the conditional probability that the ball is green, given that it is numbered 6, is 1.

(c) The ball is red, given that it has an even number.

Here, we first find the probability of both events occurring: the ball being red and having an even number. There are three red balls numbered 2, 4, and 6, and there are a total of 11 balls in the jar.

P(Red and even) = 3/11

Next, we find the probability of the event B occurring, which is the ball having an even number. There are a total of six even numbers among the balls in the jar.

P(Even) = 6/11

Finally, we can calculate the conditional probability:

P(Red|Even) = P(Red and even) / P(Even)

P(Red|Even) = (3/11) / (6/11)

P(Red|Even) = 3/6 = 1/2

So, the conditional probability that the ball is red, given that it has an even number, is 1/2.

(d) The ball has an even number, given that it is red.

Using the same approach as above, we first find the probability of both events occurring: the ball being red and having an even number. There are three red balls numbered 2, 4, and 6, and there are a total of 11 balls in the jar.

P(Red and even) = 3/11

Next, we find the probability of the event A occurring, which is the ball being red. There are a total of five red balls in the jar.

P(Red) = 5/11

Finally, we can calculate the conditional probability:

P(Even|Red) = P(Red and even) / P(Red)

P(Even|Red) = (3/11) / (5/11)

P(Even|Red) = 3/5

So, the conditional probability that the ball has an even number, given that it is red, is 3/5.

To find the conditional probabilities, we need to use the formula for conditional probability:

P(A|B) = P(A ∩ B) / P(B)

Where P(A|B) is the probability of event A occurring, given that event B has occurred, P(A ∩ B) is the probability of both events A and B occurring, and P(B) is the probability of event B occurring.

Now let's solve each part of the problem:

(a) The ball is red, given that it is numbered 2.

To find this probability, we need to count the number of red balls numbered 2 and divide it by the total number of balls numbered 2.

There is only 1 red ball numbered 2, and the total number of balls numbered 2 is 1. So the probability of drawing a red ball, given that it is numbered 2, is 1/1 = 1.

(b) The ball is green, given that it is numbered 6.

Similarly, we count the number of green balls numbered 6 and divide it by the total number of balls numbered 6.

There is only 1 green ball numbered 6, and the total number of balls numbered 6 is 1. So the probability of drawing a green ball, given that it is numbered 6, is 1/1 = 1.

(c) The ball is red, given that it has an even number.

To find this probability, we count the number of red balls with an even number and divide it by the total number of balls with an even number.

There are 2 red balls with an even number (2 and 4), and the total number of balls with an even number is 5 (2, 4, 6, 2, 4). So the probability of drawing a red ball, given that it has an even number, is 2/5.

(d) The ball has an even number, given that it is red.

To find this probability, we count the number of red balls and divide it by the total number of balls.

There are 5 red balls, and the total number of balls is 11 (5 red + 6 green). So the probability of drawing a ball with an even number, given that it is red, is 5/11.

In summary:
(a) P(Red|Numbered 2) = 1/1 = 1
(b) P(Green|Numbered 6) = 1/1 = 1
(c) P(Red|Even number) = 2/5
(d) P(Even number|Red) = 5/11

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