Question

Henry will spin the arrow on the spinner four times. What is the probability that the arrow will stop on A, then B, then C, then D?
Spinner divided into five equal parts. Parts labeled A, B, C, D, and E. Arrow pointing to B.

A. 3/5
B. 1/625
C. 1/125
D. 3/125

Answer 1

Your best answer would have to be (D.)

Answer 2

Answer: B . $\frac{1}{625}$

Step-by-step explanation:

Given: Spinner divided into five equal parts. Parts labeled A, B, C, D, and E.

Since all the five sections are distinct then, the probability that the arrow will stop on any section (A,B,C,D,E) =$\frac{1}{5}$

Now, the probability that the arrow will stop on A, then B, then C, then D [All are independent] is given by :-

$\frac{1}{5}\cdot\frac{1}{5}\cdot\frac{1}{5}\cdot\frac{1}{5}=\frac{1}{625}$

Hence, the probability that the arrow will stop on A, then B, then C, then D =$\frac{1}{625}$