Experimental probability

Of the last 80 buses to leave Nyabugogo for Musanze, 12 left late. (a) Estimate the probability that the next bus leaves late. (b) How many of the next 200 buses would you expect to leave late?

Solution

1. $P(\text{late})\approx \frac{12}{80}=0.15P(\backslash text\{late\})\; \backslash approx\; \backslash frac\{12\}\{80\}\; =\; 0.15$.
2. Expected number $=0.15\times 200=30=\; 0.15\; \backslash times\; 200\; =\; 30$ buses.

In an umudugudu, 172 of the 200 households took part in last month's Umuganda. (a) Estimate the probability that a household chosen at random took part. (b) A neighbouring umudugudu with similar habits has 350 households. About how many would you expect to take part?

Solution

1. $P(\text{took part})\approx \frac{172}{200}=0.86P(\backslash text\{took\; part\})\; \backslash approx\; \backslash frac\{172\}\{200\}\; =\; 0.86$.
2. Expected number $\approx 0.86\times 350=301\backslash approx\; 0.86\; \backslash times\; 350\; =\; 301$ households.

Ishimwe rolls a die 30 times and gets 8 sixes. Keza rolls the same die 600 times and gets 104 sixes. For a fair die, the probability of a six is $\frac{1}{6}\approx 0.167\backslash frac\{1\}\{6\}\; \backslash approx\; 0.167$. (a) Find each relative frequency. (b) Whose estimate is more reliable? (c) Does the die seem fair?

Solution

1. Ishimwe: $\frac{8}{30}\approx 0.267\backslash frac\{8\}\{30\}\; \backslash approx\; 0.267$. Keza: $\frac{104}{600}\approx 0.173\backslash frac\{104\}\{600\}\; \backslash approx\; 0.173$.
2. Keza's, because it comes from 20 times as many trials. With only 30 rolls, 8 sixes instead of the expected 5 can easily happen by chance.
3. Keza's value, 0.173, is very close to 0.167, so the die seems fair. Ishimwe's result alone might have made him wrongly think the die favours six.