Let us re-examine the salesperson problem:

• There are ten calls made in the day.
• This means that there are between 0 and 10 possible sales made in a day.
• What if we add up all of those possible probabilities?
• Let us examine this in the next example.

Example 26.1.1

Problem Setup: Again, let a salesperson call 10 clients in a day. Let the odds of the salesperson making a sale with any one of the clients be $p$=0.3

Question: What is the probability that the salesperson makes between 0 and 10 sales in a day?

Solution: In the previous section, we found the probabilities for all possible number of sales:

To calculate the probability of between 0 and 10 sales in the day, add up all these probabilities:

$\begin{align*} P(0\le x\le10) &= P(x=0)+P(x=1) + P(x=2) + P(x=3) + ... + P(x=9) + P(x=10) \\ &= 0.02825+ 0.12106+0.23347+0.26683+0.20012+0.10292+0.03676+0.009+0.00145+0.00014+0.00001 \\ &= 1 \end{align*}$

Conclusion: There is a 100% chance that the salesperson makes between 0 and 10 sales in a day. Why is this? Because that range encompasses all the possible outcomes that can happen and we know that the sum of all possible probabilities in a probability distribution must equal to 1.

• How would we do the calculation in the previous example in Excel?
• We would use Cumulative = TRUE or 1.
• Let us try this in the next example.

Example 26.1.2.

Problem Setup: We will use Excel’s binom.dist call to calculate the probability of between 0 and 10.

Solution: Click here to download the Excel solutions. Also, see the video below:

• Let us now consider the probability of at most a certain number of successes occurring.
• We denote ‘at most’ with a ‘≤’ symbol.
• It can also be stated as ‘less than or equal to.’

Example 26.2.1 – Using the Formula

Problem Setup: A hotel’s records indicate that 5% of its guests are visitors from the U.S.A.

Question: From a random sample of 12 guests, what is the probability that at most one of them is from the U.S.A.?

Solution: We know the following:

• $P(\text{at most one}) = P(x\le 1) = P(x=0) + P(x=1)$
• $n=12$ and $p=0.05$.

This gives:

• $P(x=0) = {}_{12}C_0 \cdot (0.05)^0 \cdot (1 - 0.05)^{12-0} = \frac{12!}{0!12!} \cdot (0.05)^0 \cdot (0.95)^12 =1(1)(0.5404) = 0.5404$
• $P(x=1) = {}_{12}C_1 \cdot (0.05)^1 \cdot (1 - 0.05)^{12-1} = \frac{12!}{1!11!} \cdot (0.05)^1 \cdot (0.95)^11 =12(0.05)(0.5688) = 0.34128$
• $P(x\le1) =  P(x=0)+P(x=1) = 0.5404+ 0.34128 = 0.8816$

Conclusion: There is an 88.16% chance that at most one of their guests is from the U.S.A.

In Excel’s BINOM.DIST function, when cumulative is set to TRUE (or 1):

• Excel calculates the probability of having up to and including $x$ (‘number_s‘) successes.
• BINOM.DIST($x$, $n$, $p$, 1) = P($X$ ≤ $x$)
• Let us revisit example 26.2 to better understand this concept.

Example 26.2.2 – Using BINOM.DIST()

If we wanted to solve example 26.2 using Excel’s BINOM.DIST() function:

P(x ≤ 1) = BINOM.DIST(1, 12, 0.05, 1) = BINOM.DIST(1, 12, 0.05, TRUE) = 0.8816

Click here to download the Excel solution file for the above problem.

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