Question

There was an old man with a beard

             who said, “It is just as I feared.

             Two owls and a hen,

             Four larks and a wren

             Have all built their nests in my beard”

                                                       Edgar Lear

1. A bird is chosen at random from the beard. What is the probability that it is a wren?
2. Each bird has a little tag with a letter on it, O for owl, H for hen, L for lark, and W for wren. One morning the birds are called out from the beard at random and lined up in a row from the left to the right. What is the probability that their tags are spell out WOLHOLLL?
3. How many teams of 4 birds can be chosen from the beard dwellers if each team must include at least 1 lark?
4. At least 2 larks?
5. What’s the least number of birds that can be chosen from the beard to be sure that at least 1 is a lark?
6. What’s the largest number that can be chosen without having any larks?
7. What’s the least number that can be chosen so that at least 2 birds must be of the same kind?
8. Another old man with a beard has an unlimited supply of owls, hens, larks, and wrens and wishes to choose 8 birds to nest in his beard. How many different distributions, i.e., assortments, such as 3 owls, 2 hens, no larks, and 3 wrens, can he choose?

Answer 1

Total no. of birds = 2 + 1 + 4 + 1 = 8

1) Probability that a randomly chosen bird is a wren = 1/8 = 0.125

2) P(W in the first position) = 1/8

P(O in the second position|W in the first position) = 2/7

P(Lin the third position|WO in the first two positions) = 4/6

and so on we calculate the probability for individual positions

Thus, required probability = (1/8)*(2/7)*(4/6)*(1/5)*(1/4)*1*1*1

= 1/840

3) No. of teams of 4 birds with at least 1 lark

= 4C1 * 4C3 + 4C2*4C2 + 4C3"4C1 + 4C4 = 69

4) No. of teams of 4 birds with at least 2 larks

= 4C2*4C2 + 4C3*4C1 + 4C4 = 53

5) Least number of birds that can be chosen to be sure that at least 1 is a lark = (4 + 1) = 5

(Since 4 of the birds are not lark, thus 5 birds ensure that at least 1 is a lark)

6) Largest number that can be chosen without having any larks = 4

7) The least number that can be chosen so that at least 2 birds must be of the same kind = 5

(As there is one hen, one wren and then two owls an wed four larks. Thus 5 birds ensure that at least 2 birds are present of the same kind)

8) This problem can be deduced to the problem of finding out the number of solutions of the equation

a + b + c + d = 8 where (a,b,c,d >= 0)

Thus, number of ways = = = 165

Thus, the old man can choose 165 different distributions.