Texting and Driving. In Exercises 21–26, refer to the accompan...

Question

Texting and Driving. In Exercises 21–26, refer to the accompanying table, which describes probabilities for groups of five drivers. The random variable x is the number of drivers in a group who say that they text while driving (based on data from an Arity survey of drivers).

Using Probabilities for Significant Events

a. Find the probability of getting exactly 3 drivers who say that they text while driving.

Table showing probabilities of 0 to 5 drivers texting while driving, with P(x) values from 0.066 to 0.013.

Using Probabilities for Significant Events

a. Find the probability of getting exactly 3 drivers who say that they text while driving.

Accepted Answer

Welcome back, everyone. In a poll about daily reading habits, groups of 4 individuals were asked if they read books for at least 30 minutes a day. The random variable X denotes the number of individuals in a group who say they read books for at least 30 minutes a day. Calculate the probability of exactly one individual in a group saying they read books for at least 30 minutes a day. Assume the probability that individual reads books for at least 30 minutes a day is P equals 0.5. We're given for answer choices A says 0.45. B 0.52 C 0.25, and D 0.35. For this problem, we're given the probability of an individual reading books for at least 30 minutes a day as P equals 0.5. We're going to tweet this as a binomial distribution because we have independent selections. And two possible outcomes. So now let's identify Q, which is the probability of an opposite event, finding an individual who does not read books for at least 30 minutes a day, and that's 1 minus p, which is 0.5 as well. And now our number of trials corresponds to those four individuals, right, our sample size. We have M, we have P. we have Q, and now we're interested in finding exactly one individual, right? We're interested in one success, so X equals 1. And now let's recall that according to the binomial probability formula. P X being equal to X is equal to M choose X multiplied by P raises the power of X multiplied by Q raises the power of n minus X. So in this problem we are looking for P X equals 1, and that would be N choose X. So 4, choose 1 multiplied by 0.5, that's P. raise the power of 1. Multiplied by 0.5, raise the power of n minus X, so 4 minus 1. Well done, so we have our setup, and now we can express and choose X as for factorial divided by 1 factorial multiplied by 4 minus 1 factorial. That's the formula for combinations, and we are multiplying that by 0.5. And 0.5. Raise the power of 3, right? Overall, we can just say that this is 0.5, raise the power of 4. And now if we perform the calculations, we're going to get a 0.25, which corresponds to the answer choice C. Thank you for watching.

Key Concepts

Random Variable

Probability Distribution

Calculating Probabilities

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