Questions: Based on a Comcast survey, there is a 0.8 probability that a randomly selected adult will watch prime-time TV live, instead of online, on DVR, etc. Assume that seven adults are randomly selected. Find the probability that fewer than three of the selected adults watch prime-time live. A. 0.000358 B. 0.00467 C. 0.0512 D. 0.00430

Solution Steps

We are tasked with finding the probability that fewer than three out of seven randomly selected adults watch prime-time TV live, given that the probability of an adult watching live is \( p = 0.8 \). The probability of not watching live is \( q = 1 - p = 0.2 \).

We will calculate the probabilities for exactly 0, 1, and 2 successes (i.e., adults watching live).

1. Probability of exactly 2 successes: \[ P(X = 2) = \binom{7}{2} \cdot (0.8)^2 \cdot (0.2)^{5} = 0.0043 \]

2. Probability of exactly 1 success: \[ P(X = 1) = \binom{7}{1} \cdot (0.8)^1 \cdot (0.2)^{6} = 0.00036 \]

3. Probability of exactly 0 successes: \[ P(X = 0) = \binom{7}{0} \cdot (0.8)^0 \cdot (0.2)^{7} = 1 \times 1 \times 0.0000128 = 1e-05 \]

To find the probability that fewer than three adults watch prime-time TV live, we sum the probabilities of exactly 0, 1, and 2 successes: \[ P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 1e-05 + 0.00036 + 0.0043 = 0.00467 \]

Final Answer