Question

find the indicated probabilities using the geometric distribution, the poisson distribution, or the binomial distribution. then determine if the events are unusual. if convenient, use the appropriate probability table or technology to find the probabilities.
fifty - six percent of u.s. adults oppose hydraulic fracturing (fracking) as a means of increasing the production of natural gas and oil in the united states. you randomly select six u.s. adults. find the probability that the number of u.s. adults who oppose fracking as a means of increasing the production of natural gas and oil in the united states is (a) exactly two, (b) less than four, and (c) at least three.
(a) ( p(2) = 0.176 ) (round to three decimal places as needed.)
(b) ( p(\text{less than four})=square ) (round to three decimal places as needed.)

Explanation:

Step1: Identify Distribution Type

This is a binomial probability problem. Let \( n = 6 \) (number of trials), \( p = 0.56 \) (probability of success, opposing fracking). The binomial probability formula is \( P(X = k)=\binom{n}{k}p^{k}(1 - p)^{n - k} \), and for \( P(X < 4) \), we need \( P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3) \).

Step2: Calculate \( P(X = 0) \)

\( \binom{6}{0}(0.56)^{0}(0.44)^{6}=1\times1\times0.44^{6}\approx0.000726 \)

Step3: Calculate \( P(X = 1) \)

\( \binom{6}{1}(0.56)^{1}(0.44)^{5}=6\times0.56\times0.44^{5}\approx6\times0.56\times0.00165\approx0.005544 \)

Step4: Calculate \( P(X = 2) \)

Given \( P(2)=0.176 \) (from part a, or recalculate: \( \binom{6}{2}(0.56)^{2}(0.44)^{4}=15\times0.3136\times0.00375\approx0.176 \))

Step5: Calculate \( P(X = 3) \)

\( \binom{6}{3}(0.56)^{3}(0.44)^{3}=20\times0.175616\times0.00852\approx0.302 \) (approximate, more accurately: \( 20\times0.175616\times0.085184\approx20\times0.01496\approx0.2992 \))

Step6: Sum the Probabilities

\( P(X < 4)=P(0)+P(1)+P(2)+P(3)\approx0.0007 + 0.0055+0.176+0.299\approx0.4812 \) (using more precise calculations: \( 0.000726+0.005544 + 0.176+0.2992\approx0.48147 \), rounded to three decimals: \( 0.481 \) or more accurately, let's recalculate with exact binomial coefficients:

\( \binom{6}{0}=1, \binom{6}{1}=6, \binom{6}{2}=15, \binom{6}{3}=20 \)

\( p = 0.56, 1 - p = 0.44 \)

\( P(0)=0.44^6\approx0.000726 \)

\( P(1)=6\times0.56\times0.44^5=6\times0.56\times0.00165\approx0.005544 \)

\( P(2)=15\times0.56^2\times0.44^4=15\times0.3136\times0.00375\approx0.176 \)

\( P(3)=20\times0.56^3\times0.44^3=20\times0.175616\times0.085184=20\times0.01496\approx0.2992 \)

Sum: \( 0.000726 + 0.005544 + 0.176 + 0.2992 = 0.48147 \approx 0.481 \) (rounded to three decimals)

Answer:

\( 0.481 \) (or more accurately, using calculator: \( P(X<4)=\sum_{k = 0}^{3}\binom{6}{k}(0.56)^k(0.44)^{6 - k}\approx0.481 \))