A gardener is going to plant 2 red redbushes and 2 white redbushes.If the gardener is to select each of the bushes at random,one at a time and plant them in a row, what is the probability that the 2 redbushes in the middle of the row will be redbushes. a) 1/12 b) 1/6 c) 1/5 d) 1/3 e) 1/2

htnakirs wrote:It is 1/6 Yes it is Please explain the working

The total number of ways the four bushes will be arranged = 4! = 24 Now we have W R R W as the required arrangement. The 2 red bushes can be arranged amongst themselves in 2 ways. Same holds good for the white bushes as well. Hence required probability = (2* 2) / 24 = 1/6

Can you tell me where I am wrong : Select one red from 2 => 2C1 = 2 Select second red from 1 left red redbushes =>2C2 = 1 Total ways of selection = 2*1 = 2 Now these can be arrange in 2! ways, so total ways = 2*2! = 4 similarly, the other 2 white redbushes can arrange themselves in 2! ways thus total ways = 4*2= 8 Prob = 8/24 = 1/3 Where am I wrong? Aren't we going to make a selection of 2 redbushes from the entire set?? How did you directly arrange without selecting?

gmat740 wrote:Can you tell me where I am wrong : Select one red from 2 => 2C1 = 2 Select second red from 1 left red redbushes =>2C2 = 1 Total ways of selection = 2*1 = 2 Now these can be arrange in 2! ways, so total ways = 2*2! = 4 similarly, the other 2 white redbushes can arrange themselves in 2! ways thus total ways = 4*2= 8 Prob = 8/24 = 1/3 Where am I wrong? Aren't we going to make a selection of 2 redbushes from the entire set?? How did you directly arrange without selecting? You're double counting selection and arranging, even though you mean the same thing. Let's just think this one out logically: We want W R R W There are 2 possible white bushes for first slot, 2 red bushes for 2nd slot, 1 red for 3rd slot (once we've placed one red bush) and 1 white for 4th slot (once we've placed 1st white bush), so: 2*2*1*1 = 4 desired arrangements. There are 4! = 24 total possible arrangements, so our answer is 4/24 = 1/6