Question

Assume that when adults with smartphones are randomly​ selected, 47​% use them in meetings or classes. If 10 adult smartphone users are randomly​ selected, find the probability that at least 6 of them use their smartphones in meetings or classes.

Answer 1

Answer:

$\approx$ 0.31

Step-by-step explanation:

This is a typical binomial experiment.

Let's remember what is a binomial experiment:

A binomial experiment is an experiment where you have a fixed number of independent trials with only have two outcomes (that's why is called BInomial). The probability remains constant from trial to trial.

In this case, we have 10 users (fixed number) and the result of each user doesn't affect the result of the other one (they are different people and they take their own decisions). The two outcomes are they use their smartphones in meetings or classes, or not. The probability is always the same, with a 47% of probability they use their smartphones in meetings (in our case, this is called a "success" because is what we are trying to analyze) and with a 100% - 47% = 53% they don't.

With this information, let's define X = number of users that use their smartphones in meetings or classes, X has a binomial distribution with parameters (10,47%) (10 is the total of trials and 47% the probability of success).

Let's remember that the probability of getting exactly k successes in n trials, in a binomial distribution with parameters (n, p), is given by the probability mass function:  

$P(k) = \binom{n}{k} * p^k * (1-p) ^(n-k)$

We want the probability that at least 6 of them use their smartphones, that means that we need 6, 7, 8, 9 or 10 users use them

$P(X\geq 6) = P(X=6) + P(X=7) + P(X=8) + P(X=9) + P(X=10)$

Now we just use the probability mass function, with n=10 and p=47/100 = 0.47, for each k ⊆ {6, 7, 8, 9, 10}:

$\binom{10}{6} * 0.47^6 * (1-0.47)^(10-6) + \binom{10}{7} * 0.47^7 * (1-0.47)^(10-7) + \binom{10}{8} * 0.47^8 * (1-0.47)^(10-8) + \binom{10}{9} * 0.47^9 * (1-0.47)^(10-9) + \binom{10}{10} * 0.47^(10) * (1-0.47)^(10-10) =\\ \\ \frac {10!} {6!4!} * 0.47^6 * 0.53^4 + \frac {10!} {7!3!} * 0.47^7 * 0.53^3 + \frac {10!} {8!2!} * 0.47^8 * 0.53^2 + \frac {10!} {9!1!} * 0.47^9 * 0.53^1 + \frac {10!} {10!0!} * 0.47^(10) * 0.53^0$

Now we just do the calculation and we get

$P(X\geq 6) \approx 0.31$