Assume that each bag contains 6 balls. Bag a contains 3 red and 3 white, while bag b contains 2 red, 2 white, and 2 blue. You randomly select one ball from bag a, note the color, and place the ball in bag b. You then select a ball from bag b at random and make note of its color.
(1) What is the probability that both balls are red?
(2) What is the probability that both balls are red given that the first ball you drew was red?
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To solve these questions, we can use the concept of conditional probability. Let's break down the problem step by step:
(1) What is the probability that both balls are red?
First, let's calculate the probability of selecting a red ball from bag a. Bag a contains a total of 6 balls, out of which 3 are red. Therefore, the probability of selecting a red ball from bag a is 3/6 or 1/2.
After selecting a red ball from bag a, we place it in bag b which now contains a total of 7 balls. Out of these 7 balls, 3 are red. Therefore, the probability of selecting a red ball from bag b is 3/7.
Since we need to find the probability of getting red balls from both bags, we multiply these probabilities together:
Probability of selecting a red ball from bag a * Probability of selecting a red ball from bag b = (1/2) * (3/7) = 3/14
Therefore, the probability that both balls are red is 3/14.
(2) What is the probability that both balls are red given that the first ball you drew was red?
Since we already know that the first ball selected from bag a was red, we can adjust the sample space to only consider the outcomes where the first ball is red.
Now, bag a contains 5 balls (2 red and 3 white), and bag b contains a total of 7 balls (including the red ball previously placed from bag a).
To find the probability of selecting a red ball from bag b given that the first ball was red (from bag a), we calculate the ratio of red balls in bag b to the total number of balls in bag b after adding the red ball from bag a:
Probability of selecting a red ball from bag b, given that the first ball was red = 3/7
Therefore, the probability that both balls are red given that the first ball you drew was red is 3/7.
To find the probability in both scenarios, let's break down the steps and do some calculations.
(1) Probability of both balls being red:
To solve this, we need to determine the probability of two events occurring consecutively: selecting a red ball from bag A and then selecting a red ball from bag B.
Step 1: Probability of selecting a red ball from bag A:
Since bag A contains 3 red and 3 white balls, there are 6 balls in total. Therefore, the probability of selecting a red ball from bag A is 3/6 or 1/2.
Step 2: Probability of selecting a red ball from bag B after transferring from bag A:
After transferring the ball from bag A to bag B, the contents of bag B change. Now, bag B contains 3 red, 3 white, and 2 blue balls.
The total number of balls in bag B is 8 (2 from bag A transfer + 6 initial balls in bag B). Therefore, the probability of selecting a red ball from bag B is 3/8.
To find the combined probability of both events, we multiply the probabilities:
P(both balls are red) = P(selecting red from bag A) * P(selecting red from bag B after transferring)
= (1/2) * (3/8)
= 3/16
So the probability that both balls are red is 3/16.
(2) Probability of both balls being red given that the first ball you drew was red:
In this scenario, we already know that the first ball drawn was red. Therefore, we only need to consider the probability of drawing a red ball from bag B.
Since we already transferred a red ball from bag A to bag B, there are now 2 red balls left in bag A and the total number of balls in bag B is reduced to 7.
The probability of selecting a red ball from bag B is now 2/7, as there are 2 red balls in bag B out of a total of 7 balls (2 remaining red balls from bag A transfer + 3 white balls + 2 blue balls).
Therefore, the probability of both balls being red, given that the first ball drawn was red, is 2/7.
1.
first draw is a red, now bag B contains
3R, 2W, and 2B
prob of both red = (1/2)(3/7) = 3/14
2.
prob that 2nd is read, given that the first is red = 3/7