Weekly Problem 47

9434 /www/nrich/html/content/id/9434/ 1 2012-09-18T12:42:42 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Three monkeys Barry, Harry and Larry met for tea in their favourite cafe, taking off their hats as they arrived. When they left, they each put on one of the hats at random. What is the probability that none of them left wearing the same hat as when they arrived? If you liked this problem, <a href="/4308">here is an NRICH task</a> which challenges you to use similar mathematical ideas. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Let the hats of B, H and L be b, h and l respectively. Draw a table showing the ways in which the monkeys can select hats. In only two of the six ways do none of the monkeys have the same hat as when they arrived, hence the required probability is $\frac{2}{6}=\frac{1}{3}$. Alternatively, there are $3\times2\times1=6$ possible ways to choose the three hats. There are two hats that B could choose. If B chose h, then L would have to choose b and H would have to choose l. If B chose l, then H would have to choose b and L would have to choose h. So once B has chosen his hat the other two are fixed. So there are just the two possible alternatives out of six ways. So the probability is $\frac{2}{6}=\frac{1}{3}$.</mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">Probability - Stage 3 Short Problem, UKMT 2009-2010 p24 QA4</mdoxml> 2 3 0 0 1 0 0 Weekly Problem 47 - 2012 Weekly Problem 47 - 2012 Secondary Mapping Document Probability LS Secondary Mapping Document DisplayCabinet