Question

Assume that when adults with smartphones are randomly​ selected, 51​% use them in meetings or classes. If 7 adult smartphone users are randomly​ selected, find the probability that at least 3 of them use their smartphones in meetings or classes.

Answer 1

Answer:

$P(X \geq 3) = 1-P(X$

If we find the individual probabilities we gotL

$P(X=0)=(7C0)(0.51)^0 (1-0.51)^{7-0}=0.0068$  

$P(X=1)=(7C1)(0.51)^1 (1-0.51)^{7-1}=0.0494$  

$P(X=2)=(7C2)(0.51)^2 (1-0.51)^{7-2}=0.1543$

And replacing we got:

$P(X \geq 3) = 1- [0.0068 +0.0494 +0.1543]= 0.7895$

Step-by-step explanation:

Previous concepts  

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".  

Solution to the problem  

Let X the random variable of interest, on this case we now that:  

$X \sim Binom(n=7, p=0.51)$  

The probability mass function for the Binomial distribution is given as:  

$P(X)=(nCx)(p)^x (1-p)^{n-x}$  

Where (nCx) means combinatory and it's given by this formula:  

$nCx=\frac{n!}{(n-x)! x!}$  

For this case we want to find this probability:

$P(X \geq 3)$

And we can use the complement rule for this case:

$P(X \geq 3) = 1-P(X$

If we find the individual probabilities we gotL

$P(X=0)=(7C0)(0.51)^0 (1-0.51)^{7-0}=0.0068$  

$P(X=1)=(7C1)(0.51)^1 (1-0.51)^{7-1}=0.0494$  

$P(X=2)=(7C2)(0.51)^2 (1-0.51)^{7-2}=0.1543$

And replacing we got:

$P(X \geq 3) = 1- [0.0068 +0.0494 +0.1543]= 0.7895$