If two balls are selected randomly without replacement four a bag containing 4 red and 8 green balls. List the element of the sample space corresponding probabilities.
do you want something like this ?
RR --> (4/12)(3/11) = 1/11
RG --> (4/12)(8/11) = 8/33
GR --> (8/12)(4/11) = 8/33
GG --> (8/12)(7/11) = 14/33
note that 1/11 + 8/33 + 8/33 + 14/33 = 1
To find the sample space and corresponding probabilities for selecting two balls randomly without replacement from the bag containing 4 red and 8 green balls, we can use the method of combinations.
First, let's determine the number of ways to choose 2 balls from a total of 12 (4 red + 8 green) balls:
nCr = 12C2
= 12! / (2! * (12-2)!)
= 66
Now, let's list out all the possible outcomes:
1. Selecting 2 red balls:
- RR
2. Selecting 1 red and 1 green ball:
- RG
- GR
3. Selecting 2 green balls:
- GG
The sample space consists of these possible outcomes.
Next, let's calculate the probability for each outcome:
1. Probability of selecting 2 red balls:
- P(RR) = (4/12) * (3/11) = 12/132 = 1/11
2. Probability of selecting 1 red and 1 green ball:
- P(RG) = (4/12) * (8/11) = 32/132 = 8/33
- P(GR) = (8/12) * (4/11) = 32/132 = 8/33
3. Probability of selecting 2 green balls:
- P(GG) = (8/12) * (7/11) = 56/132 = 14/33
So, the sample space corresponding to each outcome and their probabilities are:
1. RR, P(RR) = 1/11
2. RG, P(RG) = 8/33
3. GR, P(GR) = 8/33
4. GG, P(GG) = 14/33
To find the sample space and corresponding probabilities, we need to consider all the possible outcomes of selecting two balls from a bag containing 4 red and 8 green balls. Since we are selecting balls without replacement, the number of balls to choose from reduces after each selection.
Let's break it down step by step:
Step 1: Select the first ball.
- There are a total of 12 balls in the bag, so the probability of selecting a red ball on the first draw is 4/12 = 1/3, and the probability of selecting a green ball is 8/12 = 2/3.
Step 2: Select the second ball.
- After the first ball is chosen, there are only 11 balls left in the bag. The number of red and green balls remaining depends on the color of the first ball chosen.
- If the first ball chosen was red (probability 1/3):
- There are now 3 red balls and 8 green balls remaining.
- The probability of selecting a red ball on the second draw is 3/11.
- The probability of selecting a green ball on the second draw is 8/11.
- If the first ball chosen was green (probability 2/3):
- There are now 4 red balls and 7 green balls remaining.
- The probability of selecting a red ball on the second draw is 4/11.
- The probability of selecting a green ball on the second draw is 7/11.
Step 3: Combine the probabilities.
- Combine the probabilities from Step 1 and Step 2 to obtain the probabilities for each outcome in the sample space.
Here is the sample space with corresponding probabilities:
(RR) - Selecting two red balls: (1/3) * (3/11) = 1/11
(RG or GR) - Selecting one red and one green ball: (1/3) * (8/11) + (2/3) * (4/11) = 8/33 + 8/33 = 16/33
(GG) - Selecting two green balls: (2/3) * (7/11) = 14/33
Hence, the sample space corresponding to the probabilities is:
(RR) with probability 1/11
(RG or GR) with probability 16/33
(GG) with probability 14/33