Question

Suppose you roll a standard number cube and spin a spinner with four equal-sized sections labeled 1, 2, 3, 4. What is the probability you will roll a prime number and spin a prime number

Answer 1

Answer:

25% probability you will roll a prime number and spin a prime number

Step-by-step explanation:

If we have two events, A and B, and they are independent, we have that:

$P(A \cap B) = P(A) \times P(B)$

In this question:

Event A: Rolling a prime number.

Event B: Spinning a prime number.

Both the cube and the spinner have four values, ranging from one to four.

2 and 3 are prime values, that is, 2 of those values. Then

$P(A) = P(B) = \frac{2}{4} = \frac{1}{2}$

What is the probability you will roll a prime number and spin a prime number

The cube and the spinner are independent of each other. So

$P(A \cap B) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = 0.25$

25% probability you will roll a prime number and spin a prime number

Answer 2

Answer:

= 1/4

Step-by-step explanation:

In a  cube , we have  numbers labeled 1 - 6

the prime numbers we have is 2 , 3 and 5

The probability of selecting a prime number is

$=\frac{3}{6} \\\\=\frac{1}{2}$

Now this means the probability of rolling a prime number here is 1/2

Now we calculate the probability of spinning a prime number

Prime numbers here are just 2 and 3

The probability of spinning a prime number is thus 2/4 = 1/2

Thus, the probability of rolling a prime number and spinning a prime number also becomes; 1/2 * 1/2

= 1/4