51.8k views
3 votes
Nico rolled two number cubes 250 times. use the complement to find the experimantal propability of not rolling a 3 on the blue number cube, NC2.

Nico rolled two number cubes 250 times. use the complement to find the experimantal-example-1

2 Answers

4 votes

Answer:

the probability is 0%

Explanation:

is there options like

A

B

C

options

User Rescuecreative
by
7.0k points
4 votes

1.
\(P(5 \text{ on 1st cube and 2 on 2nd cube}) \approx 0.024\)

2. A.
\(P(1 \text{ on 2nd cube}) = 0.184\)

B.
\(P(\text{Not 1 on 2nd cube}) \approx 0.816\)

3.
\(P(\text{Double Sixes}) = 0.104\)

The experimental probability of double sixes is lower than expected, likely due to the limited number of trials.

1. Experimental Probability of Rolling a 5 on the First Cube and a 2 on the Second Cube:

Count the number of times the event occurs and divide it by the total number of trials.


\[ P(5 \text{ on 1st cube and 2 on 2nd cube}) = (6)/(250) \]


\[ P(5 \text{ on 1st cube and 2 on 2nd cube}) \approx 0.024 \]

2. A. Experimental Probability of Rolling a 1 on the Second Cube:


\[ P(1 \text{ on 2nd cube}) = (7 + 8 + 11 + 6 + 9 + 5)/(250) \]


\[ P(1 \text{ on 2nd cube}) = (46)/(250) \]


\[ P(1 \text{ on 2nd cube}) = 0.184 \]

B. Experimental Probability of NOT Rolling a 1 on the Second Cube:

Since there are 6 possible outcomes on a number cube, the complement of rolling a 1 is rolling any other number.


\[ P(\text{Not 1 on 2nd cube}) = 1 - P(1 \text{ on 2nd cube}) \]


\[ P(\text{Not 1 on 2nd cube}) = 1 - 0.184 \]


\[ P(\text{Not 1 on 2nd cube}) \approx 0.816 \]

3. Experimental Probability of Rolling Double Sixes:


\[ P(\text{Double Sixes}) = (26)/(250) \]


\[ P(\text{Double Sixes}) = 0.104 \]

Reasoning:

This experimental probability is lower than the theoretical probability (1/36) of rolling double sixes with two fair six-sided dice. The observed frequency might be influenced by the limited number of trials (250), and in a larger sample size, the experimental probability is likely to approach the theoretical probability.

User Okyanet
by
7.7k points