Kaushik786 wrote:
The letters of the word COMBINATION written down written in a row. Let P be the event two I's are together, Q be the event that two O's are together. Which of the following can be true ?
A. P(P) ≠ P(Q)
B. P(P) P(Q) = 2/55
C. P(P U Q) = 18/55
D. P(P) > 3/55
E. P(P ∩ Q) =1/11
There are total 11 alphabets in COMBINATION, out of which we have 2s of I, O, N.
(I) If P is the event when two I's are together and Q is the event when two O's are together, then clearly the two events are equal, as the number of Is and Os are same and they are being picked up from the same pool of alphabets.
Note: A can be eliminated straightway and B can be eliminated as the value should be a square.Total ways = \(\frac{11!}{2!2!2!}\)
(II) P is when 2 Is are together, so total alphabets are now 10 as we take Is as one alphabet.
Now, there are total 10 alphabets in COMBNAT(II)ON, out of which we have 2s of O, N.
P = \(\frac{10!}{2!2!}\)
P(P)=\(\frac{\frac{10!}{2!2!}}{\frac{11!}{2!2!2!}}=\frac{2*10!}{11!}=\frac{2}{11}\)
Note : D is correctSimilarly P(Q)=\(\frac{2}{11}\)
(III) Now P(P ∩ Q) happens when both Is and both Os are together. That is now we have 9 alphabets, where 2 Ns are together.
P ∩ Q = \(\frac{9!}{2!}\)
P(P ∩ Q)=\(\frac{\frac{9!}{2!}}{\frac{11!}{2!2!2!}}=\frac{2*2*9!}{11!}=\frac{2*2}{10*11}=\frac{2}{55}\)
Eliminate E.(IV) Let us find P(P U Q)
P(P U Q)=P(P)+P(Q)-P(P ∩ Q)=\(\frac{2}{11}+\frac{2}{11}-\frac{2}{55}=\frac{18}{55}\)
Note : C is also correctSO we have both C and D as correct option. The way it is written and for C to be the OA, the option D should be P(P) > 3/11.
C