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List IList II (A)Let n be a number chosen randomly fromthe set of first 100 natural numbers. Thenthe probability that the value of (1+i)nis real, is (P)2(B)If the coefficient of x13 in the expansion of(1x)5(1+x+x2+x3)4 is k, then thevalue of k4 is(Q)0.35(C)In an examination of 9 papers, a candidate has to pass in more papers than (s)he fails inorder to be successful. If the number ofways in which (s)he can be unsuccessful is 2m,then the value of m4 is (R)0.55(D)A,B,C are three events such that P(A)=0.6,P(B)=0.4,P(C)=0.5,P(AB)=0.8,P(AC)=0.3 and P(ABC)=0.2. If P(ABC)0.85and P(BC) lies in the interval [0.2,b],then the value of b is(S)1(T)0.5(U)0.25

Which of the following is the only INCORRECT combination?

A
(A)(U)
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B
(B)(S)
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C
(C)(R)
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D
(D)(Q)
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Solution

The correct option is C (C)(R)
(A)
We know that,
(1+i)n=(2eiπ/4)n=2n/2(cosnπ4+isinnπ4)
For (1+i)n to be real,
sinnπ4=0n=4m, mZ
So, n is a multiple of 4.
In first 100 natural numbers, there are 25 numbers that are multiples of 4.
The required probability is 25100=0.25
(A)(U)

(B)
(1x)5(1+x+x2+x3)4=(1x)5[(1+x)(1+x2)]4=(1x)(1x2)4(1+x2)4=(1x)(1x4)4=(1x)4k=0 4Ck(1)k(x4)k
So, the coefficient of x13 is,
k= 4C3(1)3=4k4=1
(B)(S)

(C)
The candidate is unsuccessful if he fails in 5 or 6 or 7 or 8 or 9 papers.
Number of ways to be unsuccessful
= 9C9+ 9C8+ 9C7+ 9C6+ 9C5= 9C0+ 9C1+ 9C2+ 9C3+ 9C4=129k=0 9Ck=12×29=28m=8m4=2
(C)(P)

(D)
P(ABC)=P(A)+P(B)+P(C)P(AB) P(BC)P(AC)+P(ABC)
We know that,
P(AB)=P(A)+P(B)P(AB)0.8=0.6+0.4P(AB)P(AB)=0.2

P(ABC)=0.6+0.4+0.50.2P(BC)0.3+0.2P(ABC)=1.2P(BC)

Now, 0.85P(ABC)1
0.851.2P(BC)10.2P(BC)0.35P(BC)[0.2,0.35]b=0.35
(D)(Q)

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