Binomial Probability Distribution

Question

Assume that when adults with smartphones are randomly​ selected, 53​% use them in meetings or classes. If 10

adult smartphone users are randomly​ selected, find the probability that fewer than 3 of them use their smartphones in meetings or classes.

Answer 1

Depending on whether or not you're allowed use of a graphing calculator, you may be expected to calculate this by hand or using a calculator. I'll present both ways of solving it here.

Solving by hand:

First, suppose I changed the question. Suppose that instead of asking "What's the probability that fewer than 3 of them use smartphones in meetings or class?" I asked "What's the probability that exactly three of them use smartphones in meetings or class?" It may seem counterintuitive to ask a different question than the one initially posed, but soon I'll show how this is relevant.

To find the probability that exactly 3 of the 10 people use their smartphones in meetings or class, we first find the probability that any specific 3 used their phones in meetings or class and that the other 7 didn't. Then we multiply by the number of combinations of 3 people that would use their smartphones in meetings or in class.

The probability that a specific 3 people all use their smartphones in meetings or class is 0.53³. The probability that the remaining 7 people do not use their smartphones in meetings or in class is then 0.47⁷. (Note that because 1-0.53 = 0.47, there is a 47% probability that a given adult does not use his or her smartphone in meetings or class.) The number of combinations of 3 people who use their smartphones in class out of a sample of 10 people is ₁₀C₃, or 10*9*8/(3*2*1) = 120.

Thus, the probability of exactly 3 people using their smartphones in meetings or in class is

₁₀C₃*0.53³*0.47⁷, or 0.0905.

In general, the probability of exactly n people using their smartphones in meetings or in class is

₁₀C_n*0.53ⁿ*0.47^10-n

How does this relate to the question as asked? The question was "What's the probability that fewer than 3 of them use smartphones in meetings or class?" Notice that this is the same as asking "What's the probability that 0 or 1 or 2 of them use smartphones in meetings or class?" We can then find the probability that 0 of them use their smartphones in meetings or in class, then the probability that 1 of them uses his or her smartphone in meetings or in class, then the probability that 2 of them use their smartphones in meetings or in class, and add them up.

If you're thinking "Wow, that sounds tedious, isn't there some faster way of doing it?" you'd be right to think that it is tedious to calculate this by hand, but unfortunately, there is no clean shortcut for finding the probability that fewer than 3 people use their smartphones in meetings or class. It's an annoying way to solve it, but there unfortunately isn't a better way. (If they had asked about the probability that fewer than 9 of them use their smartphones in meetings or class, we could calculate the opposite probability, that at least 9 of them used their smartphones in meetings or class, and subtract that from 100%)

So, adding up the 3 possibilities (that 0 of them, 1 of them, or 2 of them use their smartphones in meetings or class) we get

₁₀C₀*0.53⁰*0.47¹⁰ + ₁₀C₁*0.53¹*0.47⁹ + ₁₀C₂*0.53²*0.47⁸ = 0.0366

Solving the problem using a graphing calculator

The following instructions apply to using a TI-83 Plus calculator. The menu options may be different on other models.

Above the VARS key, in yellow, should be the word DISTR. Hit 2nd and then VARS to bring up the DISTR option. Scroll down the menu. The eleventh option should be binomcdf( which is an abbreviation for "binomial cumulative distribution function". The binomcdf function takes in 3 inputs, in the following order:

1. The number of trials, in this case, 10.
2. The probability of success on a single trial, in this case, 0.53
3. The upper bound for the range you are looking for, in this case, 2

by entering binomcdf(10, 0.53, 2) we get the answer 0.366.