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Candidates were asked to come to an interview with 3 pens each. Black, blue, green and red were the permitted pen colours that the candidate could bring. The probability that a candidate comes with all 3 pens having the same colour is _________.

3 Answers

Best answer
5 votes
5 votes

Choosing 3 pens from 4 different types of pens is same as x1 + x2 + x3 + x4 = 3 where x1,x2,x3,x4 >= 0.

So at the end we need a total of 3 pens divided into 4 groups. For dividing into 4 groups we need 3 lines.

          EX :    X | X | X |     means (1,1,1,0)  ===> Note : We have a total of 3 pens in 4 groups. 

                    X X | X | |      means (2,1,0,0)

So dividing 3 pens among 4 groups is nothing but choosing 3 lines among 6 places (3 pens + 3 lines for divding into 4 groups).

          So Total number of ways of choosing 3 pens from 4 different types of pens = C(3+3 ,3) = C(6,3) = 20.

We can choose 3 pens of same color in 4 ways (all reds , all greens , all blues , all blacks).

Probability of all 3 pens being same color = number of ways of choosing 3 pens of same color / Total number of ways of choosing 3 pens.  

                                                                   = 4 / 20

                                                                   = 0.2.

================================

Another way :

       choosing 3 pens from 4 different types can be done in 3 ways

  •  All 3 pens of same color ==> can be done in 4 ways (all reds , all greens , all blues , all blacks)
  • Two pens of same color  ==> can be done in 4 * 3 ways (4 ways of choosing 2 pens of same color and the other 1 pen can be selected in 3 ways)
  • No pens of same color   ==> can be done C(4,3) ways.

Probability of all 3 pens being same color = number of ways of choosing 3 pens of same color / Total number of ways of choosing 3 pens

                                                                  = 4 / [4 + (4*3) + C(4,3)]

                                                                  = 4 / 20

                                                                 = 0.2

selected by
3 votes
3 votes

The number of possible cases in sample space are combinations with  unlimited repitition:-

Black+Blue+Red+Green=3.

C(4-1+3,3) = 20

Now our favorable cases will be  :- 4(All red or All blue or All green or All black)

Hence required probability is :- 4/20

2 votes
2 votes
Total no. of ways of selecting 3 pens of 4 different colours:

Selecting one colour for all pens: 4C1

Selecting two colours: 4C2 x 2!

 Eg. if G and R are selected then GGR and RRG are

    2! ways

Selecting 3 colours: 4C3

So required probability is 4C1/(4C1 + 4C2 *2! + 4C3)= 4/20=0.2

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