Question

Match List-I with List-II:

List-I		List-II
(A)	If \(n=5,\ p=0.2,\ P(X=2)\)	(II)	\(0.2048\)
(B)	If \(n=7,\ p=0.6,\ P(X=3)\)	(III)	\(0.1935\)
(C)	If \(n=10,\ p=0.6,\ P(X\leq 3)\)	(IV)	\(0.0548\)
(D)	If \(n=12,\ p=0.45,\ P(4\leq X\leq 7)\)	(I)	\(0.7538\)

• A. A-III, B-I, C-IV, D-I
• B. A-I, B-II, C-III, D-IV
• C. A-II, B-III, C-IV, D-I
• D. A-II, B-III, C-I, D-IV

Hint:

For binomial distribution, use \(P(X=r)=\,^nC_rp^rq^{n-r}\), where \(q=1-p\).

Answer 1

The Correct Option is C

Concept:
For binomial distribution: \[ P(X=r)=\,^nC_rp^rq^{n-r} \] where: \[ q=1-p \]

Step 1: For A. \[ n=5,\quad p=0.2,\quad q=0.8 \] \[ P(X=2)=\,^5C_2(0.2)^2(0.8)^3 \] \[ =10(0.04)(0.512)=0.2048 \] So: \[ A\rightarrow II \]

Step 2: For B. \[ P(X=3)=\,^7C_3(0.6)^3(0.4)^4 \] \[ =35(0.216)(0.0256)=0.1935 \] So: \[ B\rightarrow III \]

Step 3: For C.
For \(n=10,\ p=0.6\): \[ P(X\leq3)=0.0548 \] So: \[ C\rightarrow IV \]

Step 4: For D.
For \(n=12,\ p=0.45\): \[ P(4\leq X\leq7)=0.7538 \] So: \[ D\rightarrow I \] \[ \therefore \text{Correct Answer is (C)} \]