Of 10 girls in a class, 3 have blue eyes. If two girls are chosen at random, what is the probability that (i) both have blue eyes, (ii) neither has blue eyes, (iii) at least one has blue eyes?
I have the answers but do not know how to get them. I'm trying to use a tree diagram but I am confusing myself.
(i) Choosing the first girl with blue eyes = 3/10. Since there is no replacement, the probability of the second girl having blue eyes = 2/9. The probability of both/all occurring is found by multiplying the individual probabilities.
(ii) Do similar process with non-blue probabilities.
(iii) This is either one or two have blue eyes. You have probability of two from (i). Probability of one = 3/10 * 7/9 or 7/10 * 3/9. (note that they have the same product). Either-or probability is found by adding the individual probabilities.
For answer (iii) are we to solve both probabilities and add them.
Ah, probability, that's always a tricky topic! Well, even Clown Bot needs some serious thinking sometimes. Let's tackle this math puzzle together, shall we?
(i) To calculate the probability that both girls have blue eyes, we need to find the probability of selecting a girl with blue eyes on the first draw and then, without replacing her, select another girl with blue eyes on the second draw.
Step 1: Probability of selecting a girl with blue eyes on the first draw = (number of girls with blue eyes) / (total number of girls) = 3/10
Step 2: Since we're not replacing the girl we picked, the probability of selecting another girl with blue eyes on the second draw is (number of girls with blue eyes left) / (total number of girls left) = 2/9
Now, to find the combined probability, we need to multiply the probabilities together:
Probability (i) = (3/10) * (2/9) = 6/90 = 1/15
(ii) The probability that neither girl has blue eyes is the opposite of both girls having blue eyes. So, we can simply subtract the probability of (i) from 1:
Probability (ii) = 1 - (1/15) = 14/15
(iii) Lastly, the probability of at least one girl having blue eyes can be found by subtracting the probability of neither girl having blue eyes from 1:
Probability (iii) = 1 - (14/15) = 1/15
There you have it! The probabilities using some math magic. Remember, math can sometimes be a bit confusing, but it's always good to approach it with a sense of humor. Hope this helps, and best of luck with your tree diagram escapades!
To solve this problem, we can use the concept of probability. Let's break it down step by step.
Step 1: Define the given information:
- Total number of girls in the class: 10
- Number of girls with blue eyes: 3
Step 2: Find the probability that both girls have blue eyes (i):
To find the probability that both girls have blue eyes, we need to calculate the probability of picking a girl with blue eyes on the first selection and then multiply it by the probability of picking another girl with blue eyes on the second selection.
P(both have blue eyes) = (number of ways to choose 2 girls with blue eyes) / (total number of ways to choose 2 girls)
The number of ways to choose 2 girls with blue eyes from a group of 3 is given by the combination formula. It can be written as C(3, 2) or 3C2, which means choosing 2 objects from a group of 3.
We can calculate this as follows:
C(3, 2) = 3! / (2!(3-2)!)
= 3! / (2!1!)
= 3
The total number of ways to choose 2 girls from a group of 10 is given by C(10, 2).
We can calculate this as follows:
C(10, 2) = 10! / (2!(10-2)!)
= 10! / (2!8!)
= (10 * 9) / (2 * 1)
= 45
So, the probability that both girls have blue eyes is:
P(both have blue eyes) = 3 / 45 = 1 / 15 ≈ 0.067
Step 3: Find the probability that neither girl has blue eyes (ii):
To find the probability that neither girl has blue eyes, we need to calculate the probability of picking a girl without blue eyes on the first selection and then multiply it by the probability of picking another girl without blue eyes on the second selection.
P(neither has blue eyes) = (number of ways to choose 2 girls without blue eyes) / (total number of ways to choose 2 girls)
The number of ways to choose 2 girls without blue eyes from a group of 7 (10 - 3) is given by C(7, 2).
We can calculate this as follows:
C(7, 2) = 7! / (2!(7-2)!)
= 7! / (2!5!)
= (7 * 6) / (2 * 1)
= 21
The total number of ways to choose 2 girls from a group of 10 is given by C(10, 2), which we calculated previously as 45.
So, the probability that neither girl has blue eyes is:
P(neither has blue eyes) = 21 / 45 = 7 / 15 ≈ 0.467
Step 4: Find the probability that at least one girl has blue eyes (iii):
To find the probability that at least one girl has blue eyes, we need to subtract the probability that neither girl has blue eyes from 1.
P(at least one has blue eyes) = 1 - P(neither has blue eyes)
We already calculated the probability that neither girl has blue eyes as 7 / 15.
So, the probability that at least one girl has blue eyes is:
P(at least one has blue eyes) = 1 - 7 / 15 = 8 / 15 ≈ 0.533
Therefore, the answers are:
(i) The probability that both have blue eyes is approximately 0.067
(ii) The probability that neither has blue eyes is approximately 0.467
(iii) The probability that at least one has blue eyes is approximately 0.533
To solve this problem, you can use the concept of probability. Let's break down each part of the question:
(i) To find the probability that both girls have blue eyes, you need to determine the probability of one girl having blue eyes, and then multiply it by the probability of the second girl having blue eyes.
First, find the probability of one girl having blue eyes:
Given that there are 10 girls in total and 3 of them have blue eyes, the probability of selecting a girl with blue eyes is 3/10.
Next, since the first girl with blue eyes has already been chosen, there are now 2 girls with blue eyes out of the remaining 9 total girls. Therefore, the probability of selecting a second girl with blue eyes is 2/9.
To find the probability that both girls have blue eyes, multiply these two probabilities together: (3/10) * (2/9) = 6/90 = 1/15.
(ii) To find the probability that neither girl has blue eyes, consider the complement of having at least one girl with blue eyes.
Given that there are 10 girls in total and 3 of them have blue eyes, the probability of selecting a girl without blue eyes is 7/10.
Since there is no replacement, after selecting the first girl without blue eyes, there are now 6 girls without blue eyes out of the remaining 9 total girls. Thus, the probability of selecting a second girl without blue eyes is 6/9.
To find the probability that neither girl has blue eyes, multiply these two probabilities together: (7/10) * (6/9) = 42/90 = 7/15.
(iii) To find the probability that at least one girl has blue eyes, subtract the probability of neither girl having blue eyes from 1.
The probability of neither girl having blue eyes, as calculated in part (ii), is 7/15. Subtracting this from 1 gives: 1 - (7/15) = 8/15.
So, the probability that at least one girl has blue eyes is 8/15.
By following these steps and understanding the concepts of probability, you can arrive at the correct answers.