Probability sampling involving holidays

Question

Four people choose at random one holiday destination among ten destinations. Show that the probability that at least two people will make the same choice is less than 1/2.

So far I have deduced that sampling is unordered and repetition is allowed, so to find the size of the sample space, I did (n+r-1)C(r), which gave me 715. Then to find the size of the event two people choose the same destination, I did 1/10*1/10, and then divided these two sizes to get 1/71500.

Is this the correct answer?

So 1/71500 is clearly less than 1/2

Answer 1

You are going about this the wrong way. The probability that at least two people select the same destination is equal to 1 minus the probability that all four people choose a different destination. In the latter case, the first person can select 10 destinations, the second person 9 (as he or she cannot select the one chosen by the first person), the third person 8 and the fourth person 7. The probability that at least two people select the same destination thus equals:

$$1 - \frac{10}{10} \cdot \frac{9}{10} \cdot \frac{8}{10} \cdot \frac{7}{10} = 0.496 < 0.5$$

Answer 2

number of all possible holiday choices = $10^4$ number of holiday choices with multiple bookings not possible = 10x9x8x7 = 10!/ 6!

P(no multiple booking) = $(10! / 6!) / 10^4 = \frac{63}{125}$

there is therefore a $\frac{62}{125}$ chance that there will be a duplicate choice