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Today Puzzle #525

Puzzle No. 525 鈥� Wednesday 16 July

Jane owns three hens: Tilly, Olivia and Gillian. Each week Tilly lays 1 egg, Olivia lays 2 eggs and Gillian lays 3 eggs. At the end of the week Jane gathers up all six eggs and randomly selects three of them to cook herself an omelette. What is the probability that each hen has contributed to the omelette?

Today’s #PuzzleForToday has been set by Dr Geoff Evatt, School of Mathematics, University of Manchester

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30%. There are 6 ways in which eggs can be selected so that each of the three hens has contributed an egg to the omelette. Denoting the hens by T, O and G, the possible combinations for selecting the eggs are the orderings: TOG, TGO, OTG, OGT, GTO, GOT. Taking the first possibility (TOG), it can be calculated by noting there is a 1 in 6 chance of selecting Tilly's egg first, then a 2 in 5 probability of selecting one of Olivier's two eggs from the remaining five eggs, and then a 3 in 4 probability of selecting one of Gillian's three eggs from the remaining four eggs. And so the probability of that combination is: 1/6 x 2/5 x 3/4 =6/120 = 1/20 = 5%. Fortunately calculating the probability of any one of these 6 events is the same, (e.g. GOT is 3/6 x 2/5 x 1/4 = 5%). And so the probability of selecting three eggs each laid by a different hen, is 6 times 5% =30%.

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