Question

A 2005 Gallup Poll found that 7% of teenagers (ages 13 to 17) 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. b) Calculate the probability that exactly 2 of them suffer from arachnophobia. c) Calculate the probability that at most 1 of them suffers from arachnophobia. 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 the counselor to randomly assign teenagers to tents?

Answer 1

Answer:

Step-by-step explanation:

A. The probability that at least one of them suffers from arachnophobia

Let x be the number of students who suffer from arachnophobia.

                = P(x ≥ 1)

                = 1 - P(x = 0)

               = 1 - $\begin{bmatrix}10\\0 \end{bmatrix}$ 0.07⁰(1 - 0.07)¹⁰⁻⁰

               = 1 - (0.93)¹⁰

               = 0.516

B.  The probability that exactly 2 of them suffer from arachnophobia.

              =  P(x = 2)

              = $^{10}c_2$ (0.07)² (0.93)⁸

              = 0.1234

C.  The probability that at most 1 of them suffers from arachnophobia.

      P(x ≤ 1)

       = P(x = 0) + P(x = 1)

       = 0.4848 + 0.3643

      = 0.8483

D. Yes, it seems reasonable for the counselor to randomly assign teenagers to tents because the probability of arachnophobia in students is very less.

Answer 2

A) The probability that at least one of them suffers from arachnophobia is; P(x ≥ 1) = 0.51602

B) The the probability that exactly 2 of them suffer from arachnophobia is; P(x = 2) = 0.1234

C) The probability that at most 1 of them suffers from arachnophobia is; P(x ≤ 1) = 0.84826

D) Yes it will be reasonable.

This is a binomial probability distribution question with the formula;

P(X = r) = nCr × p^(r) × q^(n - r)

We are given;

p = 7% = 0.07

Thus; q = 1 - 0.07

q = 0.93

n = 10

A) Probability that at least one of them suffers from arachnophobia will be;

P(x ≥ 1) = 1 - P(x = 0)

Now;

P(x = 0) = 10C0 × 0.07⁰ × (0.93)¹⁰⁻⁰

P(x = 0) = 0.48398

Thus;

P(x ≥ 1) = 1 - 0.48398

P(x ≥ 1) = 0.51602

B) The probability that exactly 2 of them suffer from arachnophobia is given by;

P(x = 2) = 10C2 × (0.07)² × (0.93)⁸

P(x = 2) = 45 × 0.0049 × 0.5596

P(x = 2) = 0.1234

C) The probability that at most 1 of them suffers from arachnophobia is;

P(x ≤ 1) = P(x = 0) + P(x = 1)

P(x = 1) = 10C1 × (0.07)¹ × (0.93)^(9)

P(x = 1) = 10 × 0.07 × 0.5204

P(x = 1) = 0.36428

From earlier, we saw that P(x = 0) = 0.48398.

Thus;

P(x ≤ 1) = 0.48398 + 0.36428

P(x ≤ 1) = 0.84826

D) Yes, it would seem reasonable for the counselor to randomly assign teenagers to tents due to the fact that the probability of teenagers suffering from arachnophobia is very small at 7%.

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