birds comb qs
6 birds, all of a different kind (X1, X2, . ..X6) are put in pairs 3 cages. However, the pairs (X1, X2), (X2, X3). . . .(X5, X6) cannot be put together in a cage.
Q1. All the three pairs cannot be determined if it is known that X1 is put together in a cage with
a. X4 b. X5 c. X3 d. X6
Q2. If one of the pairs in a cage is given, then what is the probability that all the other required pairs can be determined?
a. 2/5 b. 1/2 c. 5/11 d. 3/5
X1 CAN BE PAIRED WITH 3,4,5,6(4 WAYS). X2 WITH 4,5,6(3WAYS). X3 WITH 1,5,6(3 WAYS). X4 WITH 1,2,6(3 WAYS). X5 WITH 1,2,3(3 WAYS). AND X6 WITH 1,2,3,4(4 WAYS).
IF X1 IS PAIRED WITH X3 THEN X5 IS LEFT WITH X2 ONLY.REST CAN BE DETERMINED.
IF X1 IS PAIRED WITH X5(or X6) THEN X3(X4 in case of X6) IS LEFT WITH X6(X2 in case of X6.) ONLY.REST AGAIN CAN BE DETERMINED.
IF X1 IS PAIRED WITH X4 THEN X2 CAN BE PAIRED WITH 5,6. X3 WITH 5,6. X5 WITH 2,3. X6 WITH 2,3. SO IN THIS CASE CANT BE DETERMINED.HENCE X4 IS THE ANSWER OF FIRST QUESTION.
SECOND QUESTION: IF X1 IS PAIRED WITH 3,5,6 PAIRS CAN BE DETERMINED. TOTAL 3 WAYS.
IF X2 IS PAIREDWITH X6 PAIRS CAN BE DETERMINED.1 WAY
X3 WITH X1.1 WAY: X4 WITH X6 1 WAY: X5 WITH X1.1 WAY AND X6 WITH 1,2,4 .TOTAL 3 WAYS.ALL WAYS ARE ADDED U GET 10 WAYS.
TOTAL POSSIBLE(SEE IN BOLD LETTERS ABOVE) = 4+3+3+3+3+4=20.
SO THE PROBABILITY FOR THE SECOND QUESTION IS 10/20 =1/2.
Answers are
Q1. a. X4
Q2 b. 1/2
Please verify it.