Let us revisit the hotel example from the previous section to highlight the similarities and differences between ‘more than’ and ‘at least’ calculations.

Example 28.1 – Using the Formula

Problem Setup: A hotel’s records indicate that 65% of its guests are visitors from Canada.

Question: From a random sample of 12 guests, what is the probability that at least 10 of them are from Canada?

Solution: We know the following:

• $P(\text{at least 10}) = P(x\ge 10) = P(x=10)+P(x=11) + P(x=12)$
• $n=12$ and $p=0.65$.

This gives:

• $P(x=10) = {}_{12}C_{10} \cdot (0.65)^{10} \cdot (1 - 0.65)^{12-10} = \frac{12!}{10!2!} \cdot (0.65)^{10} \cdot (0.35)^2 =66(0.01346)(0.1225) = 0.10885$
• $P(x=11) = {}_{12}C_{11} \cdot (0.65)^{11} \cdot (1 - 0.65)^{12-11} = \frac{12!}{11!1!} \cdot (0.65)^{11} \cdot (0.35)^1 =12(0.00875)(0.35) = 0.03675$
• $P(x=12) = {}_{12}C_{12} \cdot (0.65)^{12} \cdot (1 - 0.65)^{12-12} = \frac{12!}{12!0!} \cdot (0.65)^{12} \cdot (0.35)^0 =1(0.00569)(1) = 0.00569$
• $P(x\ge 10) =  P(x=10)+P(x=11)+P(x=12) = 0.10885+0.03675+ 0.00569 = 0.1513$

Conclusion: There is an 15.13% chance that at least 10 of them are from Canada.

• Using Excel’s BINOM.DIST() function is much quicker than using the formula shown in the previous section
• Again, just be careful of which $x$ value to include and don’t forget to take the complement.
• We will try this out in the next example.

Example 28.2 – Using Excel and a Complement

Problem Setup: Let us now revisit example 28.1 but we will use Excel.

Question: Can you use Excel’s BINOM.DIST to calculate the probability of at least 10 guests being from Canada?

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

Conclusion: Again, here is an 15.13% chance that at least 10 of them are from Canada.

• Are there any notes you want to take from this section? Is there anything you’d like to copy and paste below?
• These notes are for you only (they will not be stored anywhere)
• Make sure to download them at the end to use as a reference