Target Exam

CUET

Subject

-- Applied Mathematics - Section B2

Chapter

Probability Distributions

Question:

A coin is tossed until a head appears or tail appears 4 times in succession Match P.D. of x.

Match List-I with List-II. (x is number of times the coin is tossed)

List-I List-II
(A) 1 (I) $\frac{1}{4}$
(B) 2 (II) $\frac{1}{2}$
(C) 3 (III) $\frac{1}{8}$
(D) 4 (IV) $\frac{1}{8}$

Choose the correct answer from the options given below :

Options:

(A)-(I), (B-(II), (C)-(III), (D)-(IV)

(A)-(II), (B-(I), (C)-(III), (D)-(IV)

(A)-(II), (B-(II), (C)-(I), (D)-(IV)

(A)-(II), (B-(I), (C)-(IV), (D)-(III)

Correct Answer:

(A)-(II), (B-(I), (C)-(III), (D)-(IV)

Explanation:

The correct answer is Option (2) → (A)-(II), (B-(I), (C)-(III), (D)-(IV)

$P(X=1)=P(H)=\frac{1}{2}$

$P(X=2)=P(T)×P(H)=\frac{1}{2}×\frac{1}{2}=\frac{1}{4}$

$P(X=3)=P(T)×P(T)×P(H)=\frac{1}{2}×\frac{1}{2}×\frac{1}{2}=\frac{1}{8}$

$P(X=4)=P(T)×P(T)×P(T)×(P(T)×P(H))=\frac{1}{2}×\frac{1}{2}×\frac{1}{2}×(\frac{1}{2}×\frac{1}{2})=\frac{1}{8}$