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 - two 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 seven 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 three, (b) less than four, and (c) at least three.
(a) ( p(3) = 0.261 ) (round to three decimal places as needed.)
(b) ( p(\text{less than four}) = 0.456 ) (round to three decimal places as needed.)
(c) ( p(\text{at least three}) = square ) (round to three decimal places as needed.)

Explanation:

Step1: Identify Distribution Type

This is a binomial distribution problem with \( n = 7 \) (number of trials), \( p = 0.52 \) (probability of success, since 52% oppose), and we need \( P(X \geq 3) \). The formula for binomial probability is \( P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k} \), and \( P(X \geq 3)=1 - P(X < 3)=1 - [P(X = 0)+P(X = 1)+P(X = 2)] \).

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

\( \binom{7}{0} (0.52)^0 (0.48)^7 = 1 \times 1 \times 0.48^7 \approx 0.0059 \)

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

\( \binom{7}{1} (0.52)^1 (0.48)^6 = 7 \times 0.52 \times 0.48^6 \approx 7 \times 0.52 \times 0.0123 \approx 0.0444 \)

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

\( \binom{7}{2} (0.52)^2 (0.48)^5 = 21 \times 0.2704 \times 0.0256 \approx 21 \times 0.00692 \approx 0.1453 \)

Step5: Sum \( P(X < 3) \)

\( P(X < 3)=0.0059 + 0.0444 + 0.1453 = 0.1956 \)

Step6: Calculate \( P(X \geq 3) \)

\( P(X \geq 3)=1 - 0.1956 = 0.8044 \) (rounded to four decimals, adjust to three: 0.804) Wait, recalculate more accurately.

Wait, let's use a calculator for binomial probabilities:

\( P(X=0) = 0.48^7 \approx 0.005897 \)

\( P(X=1) = 7 \times 0.52 \times 0.48^6 = 7 \times 0.52 \times 0.012285 \approx 7 \times 0.006388 \approx 0.044716 \)

\( P(X=2) = 21 \times 0.52^2 \times 0.48^5 = 21 \times 0.2704 \times 0.025594 \approx 21 \times 0.006929 \approx 0.145509 \)

Sum: \( 0.005897 + 0.044716 + 0.145509 = 0.196122 \)

Then \( P(X \geq 3) = 1 - 0.196122 = 0.803878 \approx 0.804 \) (rounded to three decimal places). Wait, but maybe using binomial table or more precise calculation. Alternatively, check with another method. Wait, maybe I made a mistake in \( p \): 52% is 0.52, correct. \( n=7 \).

Wait, let's use the binomial formula correctly:

\( \binom{7}{k} = \frac{7!}{k!(7 - k)!} \)

For \( k=0 \): \( 1 \), \( k=1 \): \( 7 \), \( k=2 \): \( 21 \), \( k=3 \): \( 35 \), etc.

Recalculate \( P(X=2) \): \( 0.52^2 = 0.2704 \), \( 0.48^5 = 0.48 \times 0.48 \times 0.48 \times 0.48 \times 0.48 = (0.48^2)^2 \times 0.48 = 0.2304^2 \times 0.48 = 0.05308416 \times 0.48 = 0.0254803968 \). Then \( 21 \times 0.2704 \times 0.0254803968 = 21 \times 0.006890 = 0.14469 \).

\( P(X=1) \): \( 7 \times 0.52 \times 0.48^6 \). \( 0.48^6 = 0.48^5 \times 0.48 = 0.0254803968 \times 0.48 = 0.01223059046 \). Then \( 7 \times 0.52 \times 0.01223059046 = 7 \times 0.00635990704 = 0.0445193493 \).

\( P(X=0) \): \( 0.48^7 = 0.48^6 \times 0.48 = 0.01223059046 \times 0.48 = 0.00587068342 \).

Sum \( P(X < 3) = 0.00587068342 + 0.0445193493 + 0.14469 = 0.19508 \). Then \( 1 - 0.19508 = 0.80492 \approx 0.805 \) (rounded to three decimal places). Wait, maybe the initial a and b were calculated with binomial, so let's check with binomial calculator.

Using binomial probability formula for \( n=7 \), \( p=0.52 \):

\( P(3) = \binom{7}{3} 0.52^3 0.48^4 = 35 \times 0.140608 \times 0.05308416 = 35 \times 0.007465 = 0.261275 \) (matches part a: 0.261).

\( P(less than 4) = P(0)+P(1)+P(2)+P(3) = 0.00587 + 0.04452 + 0.14469 + 0.26128 = 0.45636 \) (matches part b: 0.456).

Then \( P(at least 3) = P(3)+P(4)+P(5)+P(6)+P(7) \).

Calculate \( P(4) = \binom{7}{4} 0.52^4 0.48^3 = 35 \times 0.07311616 \times 0.110592 = 35 \times 0.008086 = 0.28301 \)

\( P(5) = \binom{7}{5} 0.52^5 0.48^2 = 21 \times 0.0380204032 \times 0.2304 = 21 \times 0.008760 = 0.18396 \)

\( P(6) = \binom{7}{6} 0.52^6 0.48^1 = 7 \times 0.01977060966 \times 0.48 = 7 \times 0.00949 \approx 0.06643 \)

\( P(7) = \binom{7}{7} 0.52^7 0.48^0 = 1 \times 0.01028071702 \times 1 = 0.01028 \)

Now sum \( P(3) + P(4) + P(5) + P(6) + P(7) = 0.26128 + 0.28301 + 0.18396 + 0.06643 + 0.01028 = 0.80496 \approx 0.805 \) (rounded to three decimal places).

Answer:

0.805