Five Cub Scouts are in a den; A, B, C, D and E. A has a Bobcat Badge, B has a Wolf Badge, C has a Bear Badge, D is a slacker that only has a bunch of Gold Arrows, and E wishes that he was a Girl Scout. If two are selected from this group to sell a fund raising item, what are the odds that the group will include one scout that has an animal based badge, and that the group will end up selling cookies? A: 2/10 B: 2/5 C: 19/20 D: 3/20 E: 5/2

The question is a lot american: My assumptions only girl scout can sell cookies Number of ways of selecting 2 in 5 = 5C2 One animal badge and 1 girl scout = 3C1+1 : D is named a slacker so 4/10 = 2/5 .. now we need to find the odds against this IMO E. ( I am having very little confidence on my solution)

I totally agree with HSPA , going ahead with same assumption . IMO D. (3/5) X (1/4) = 3/20 3 ppl with animal badge so 3/5 1 person wanna be girl scout so 1/4 [1 person is already selected so 4 left]

edvhou812 wrote:Five Cub Scouts are in a den; A, B, C, D and E. A has a Bobcat Badge, B has a Wolf Badge, C has a Bear Badge, D is a slacker that only has a bunch of Gold Arrows, and E wishes that he was a Girl Scout. If two are selected from this group to sell a fund raising item, what are the odds that the group will include one scout that has an animal based badge, and that the group will end up selling cookies? A: 2/10 B: 2/5 C: 19/20 D: 3/20 E: 5/2 out of five scouts any two scouts can be selected in 5C2 ways. now as the group must include one animal based badge so out of 3 people the one with animal badge be selected in 3C1,as no restriction is placed upon the second selection therefore out of the remaining two persons one can be selected in 2C1 ways. therefore required probability would be 1 - P(favour) = 1- 3C1*2C1/5C2; 1-3/5=2/5 hence B