A 2005 Gallup Poll found that 7% of teenagers suffer from arachnophobia and are extremely afraid of spiders. At a summer camp there are 10 teenagers sleeping in each tent. Assume that these 10 teenagers are independent of each other.
a) Calculate the probability that at least one of them suffers from arachnophobia.
Is it 0.516?
b) Calculate the probability that exactly 2 of them suffer from arachnophobia.
Is it 0.123
c) Calculate the probability that at most 1 of them suffers from arachnophobia.
Is it 0.848?
d)If the camp counselor wants to make sure no more than 1 teenager in each tent is afraid of spiders, does it seem reasonable for him to randomly assign teenagers to tents?
Is it 1? I am not sure about this.
Yes, it seems reasonable for the camp counselor to randomly assign teenagers to tents. The probability that at most one of them suffers from arachnophobia is 0.848, which is quite high.
To solve these probability problems, we can use the concept of binomial distribution since each teenager's fear of spiders can be viewed as a success or failure with a fixed probability.
a) The probability that at least one teenager suffers from arachnophobia can be calculated by finding the complement of the probability that none of them suffer from it.
P(at least one) = 1 - P(none)
To compute P(none), we can use the probability of not having arachnophobia, which is 1 - 0.07 = 0.93.
P(at least one) = 1 - P(none) = 1 - (0.93)^10 ≈ 0.516
Therefore, the probability that at least one teenager suffers from arachnophobia is approximately 0.516.
b) To find the probability that exactly 2 of them suffer from arachnophobia, we can use the binomial probability formula:
P(exactly k successes) = (nCk) * p^k * (1-p)^(n-k)
Here, n represents the number of trials (10), k represents the number of successes (2), and p represents the probability of success (0.07).
P(exactly 2) = (10C2) * (0.07)^2 * (1-0.07)^(10-2) ≈ 0.147
So, the probability that exactly 2 teenagers suffer from arachnophobia is approximately 0.147.
c) To calculate the probability that at most 1 of them suffers from arachnophobia, we need to find the sum of probabilities of 0 and 1 teenager having arachnophobia.
P(at most 1) = P(0) + P(1)
P(0) = (10C0) * (0.07)^0 * (1-0.07)^(10-0) = (1) * (1) * (0.93)^10 ≈ 0.394
P(1) = (10C1) * (0.07)^1 * (1-0.07)^(10-1) = (10) * (0.07) * (0.93)^9 ≈ 0.454
P(at most 1) = P(0) + P(1) ≈ 0.394 + 0.454 ≈ 0.848
Therefore, the probability that at most 1 teenager suffers from arachnophobia is approximately 0.848.
d) If the camp counselor wants to ensure that no more than 1 teenager in each tent is afraid of spiders, it may not be reasonable to randomly assign teenagers to tents.
Since the probabilities of teenagers having arachnophobia are independent, random assignment may lead to some tents having more than one teenager with arachnophobia. To guarantee that no more than one teenager in each tent is afraid of spiders, the counselor may need to consider the distribution of teenagers with arachnophobia and their grouping within tents more purposefully.
To calculate the probabilities, we will use the binomial distribution formula:
P(X = k) = (nCk) * p^k * (1-p)^(n-k)
Where:
- n is the number of trials or teenagers (10 in this case)
- k is the number of successes or teenagers suffering from arachnophobia
- p is the probability of success or the probability that a given teenager suffers from arachnophobia
- (nCk) is the number of combinations of n objects taken k at a time
a) To find the probability that at least one of the ten teenagers suffers from arachnophobia, we need to calculate the probability that none of them suffer from it and subtract it from 1.
P(at least one suffers) = 1 - P(none suffers)
P(none suffers) = (1 - p)^n
= (1 - 0.07)^10
P(at least one suffers) = 1 - (1 - 0.07)^10
= 1 - 0.64009
= 0.35991
Therefore, the probability that at least one of them suffers from arachnophobia is approximately 0.36, not 0.516.
b) To find the probability that exactly 2 of the ten teenagers suffer from arachnophobia, we can directly apply the binomial distribution formula with n = 10, k = 2, and p = 0.07.
P(X = 2) = (10C2) * 0.07^2 * (1 - 0.07)^(10 - 2)
P(X = 2) = (45) * 0.07^2 * (0.93)^8
≈ 0.123
Therefore, the probability that exactly 2 of the teenagers suffer from arachnophobia is approximately 0.123.
c) To find the probability that at most 1 of the ten teenagers suffer from arachnophobia, we need to calculate the probability that either none or only one of them suffers from it.
P(at most 1 suffers) = P(none suffers) + P(one suffers)
P(none suffers) = (1 - p)^n
= (1 - 0.07)^10
P(one suffers) = (nC1) * p^1 * (1 - p)^(n - 1)
= (10C1) * 0.07^1 * (1 - 0.07)^(10 - 1)
P(at most 1 suffers) = (1 - p)^n + (nC1) * p^1 * (1 - p)^(n - 1)
≈ (1 - 0.07)^10 + (10C1) * 0.07^1 * (1 - 0.07)^(10 - 1)
≈ 0.848
Therefore, the probability that at most 1 of the ten teenagers suffer from arachnophobia is approximately 0.848.
d) From the calculations above, we see that the probability of having at most 1 teenager suffering from arachnophobia in each tent is approximately 0.848. Since this probability is high, it seems reasonable for the camp counselor to randomly assign teenagers to tents to ensure no more than 1 teenager in each tent is afraid of spiders. However, other factors like tent sizes, gender distribution, or other considerations might also affect the decision-making process.