Question

A spinner has 10 equally sized sections, 4 of which are gray and 6 of which are blue. The spinner is spun twice. What is the probability that the first spin lands on blue and the second spin lands on gray.

Answer 1

The probability of that is 6/25

Answer 2

Answer:  0.24

Step-by-step explanation:

We use the formula ,

$\text{Probability}=\dfrac{\text{favorable outcomes}}{\text{total outcomes}}$

Given : A spinner has 10 equally sized sections, 4 of which are gray and 6 of which are blue.

Event 1 : spin lands on blue

$P(E_1)=\dfrac{6}{10}$

Event 2 : spin lands on gray

$P(E_2)=\dfrac{4}{10}$

Since both the events are independent of each other , then the probability that the first spin lands on blue and the second spin lands on gray will be :-

$P(E_1)\times P(E_2)\\\\=\dfrac{6}{10}\times\dfrac{4}{10}=0.24$

Hence, the probability that the first spin lands on blue and the second spin lands on gray = 0.24