Question

enter the answer in each space provided. use numbers instead of words.

$p(1 \text{ on 1st spinner}) = \frac{1}{6}$

$p(1 \text{ on 2nd spinner}) = \frac{1}{4}$

$p(\text{any number on 3rd spinner}) = 1$

$p(1 \text{ on 1st spinner and } 1 \text{ on 2nd spinner}) = \square$

Explanation:

Step1: Identify the rule for independent events

For two independent events \( A \) and \( B \), the probability of both occurring is \( P(A \text{ and } B) = P(A) \times P(B) \). Here, \( A \) is "1 on 1st spinner" and \( B \) is "1 on 2nd spinner".

Step2: Substitute the known probabilities

We know \( P(1 \text{ on 1st spinner}) = \frac{1}{6} \) and \( P(1 \text{ on 2nd spinner}) = \frac{1}{4} \). So we calculate the product: \( \frac{1}{6} \times \frac{1}{4} \).

Step3: Calculate the product

\( \frac{1}{6} \times \frac{1}{4} = \frac{1 \times 1}{6 \times 4} = \frac{1}{24} \).

Answer:

\(\frac{1}{24}\)