This is a two step experiment where each step is independent of the other. The probability of both succeed is the product of the probability of each step. [called the product rule]
Step 1:
P(even)=P({2,4,6} out of {1,2,3,4,5,6})=3/6
P(2)=P({2} out of {1,2,3,4,5,6})=1/6
Now use the product rule to find the probability of both events happening (success), simplify the fraction as requested.
Note: in general, it is preferable to use fractions for probability problems because answers are exact. Using the calculator will give inaccurate or ambiguous results due to round off errors.
You roll a number cube twice. Find P(even, then not 2). Write the probability as a fraction in simplest form. Explain your answer. Please help!
15 answers
It said NOT 2; therefore, the product of 1/2 times 5/6 would be your answer
so 5/12?
not me and my grandma tryin find these answers ✋
can you please explain?
well if i prank i guess im not a fully good person but im nice :)
You're still not a good person if you call people a loser.
well not if i dont mean it LOL
ummm is anyone going to answer my question from literally 2017? My little sister needs help but I dont know how to explain to her.
Srsly?? We still haven't answered this question from 2017?!??
I apologize for the delay in answering your question. Here is the solution to your problem:
To find the probability of rolling an even number on the first roll and not rolling a 2 on the second roll, we need to find the probability of each event happening separately and then multiply them.
The probability of rolling an even number on the first roll is 3/6 (since there are three even numbers in a six-sided die). The probability of not rolling a 2 on the second roll is 5/6 (since there are five numbers that are not 2 in a six-sided die).
Therefore, the probability of rolling an even number on the first roll and not rolling a 2 on the second roll is:
P(even, then not 2) = P(even) × P(not 2) = (3/6) × (5/6) = 15/36 = 5/12
So the probability of rolling an even number on the first roll and not rolling a 2 on the second roll is 5/12.
To find the probability of rolling an even number on the first roll and not rolling a 2 on the second roll, we need to find the probability of each event happening separately and then multiply them.
The probability of rolling an even number on the first roll is 3/6 (since there are three even numbers in a six-sided die). The probability of not rolling a 2 on the second roll is 5/6 (since there are five numbers that are not 2 in a six-sided die).
Therefore, the probability of rolling an even number on the first roll and not rolling a 2 on the second roll is:
P(even, then not 2) = P(even) × P(not 2) = (3/6) × (5/6) = 15/36 = 5/12
So the probability of rolling an even number on the first roll and not rolling a 2 on the second roll is 5/12.
i need a simplifyed answer for this cause im stupid
No problem! Let me simplify the answer for you.
To find P(even, then not 2), we need to find the probability of two events happening one after the other: first rolling an even number, and then not rolling a 2.
The probability of the first event happening (rolling an even number) is 3/6, which simplifies to 1/2.
The probability of the second event happening (not rolling a 2) is 5/6.
To find the probability of both events happening, we multiply the probabilities:
P(even, then not 2) = (1/2) x (5/6) = 5/12
So the probability of rolling an even number on the first roll and not rolling a 2 on the second roll is 5/12.
To find P(even, then not 2), we need to find the probability of two events happening one after the other: first rolling an even number, and then not rolling a 2.
The probability of the first event happening (rolling an even number) is 3/6, which simplifies to 1/2.
The probability of the second event happening (not rolling a 2) is 5/6.
To find the probability of both events happening, we multiply the probabilities:
P(even, then not 2) = (1/2) x (5/6) = 5/12
So the probability of rolling an even number on the first roll and not rolling a 2 on the second roll is 5/12.
thats the same thing bot
I apologize, you are correct. My previous response included a simplified answer as well. Here is another explanation, using different language:
When you roll a number cube twice, there are 36 possible outcomes (6 possible outcomes on the first roll, and 6 possible outcomes on the second roll).
We want to find the probability of two specific events happening in this order: first rolling an even number, and then not rolling a 2.
There are three even numbers on a number cube: 2, 4, and 6. So the probability of rolling an even number on the first roll is 3 out of 6, or 1/2.
There are five numbers that are not 2: 1, 3, 4, 5, and 6. So the probability of not rolling a 2 on the second roll is 5 out of 6.
To find the probability of both events happening, we multiply the probabilities:
P(even, then not 2) = (1/2) x (5/6) = 5/12
So the probability of rolling an even number on the first roll and not rolling a 2 on the second roll is 5/12.
When you roll a number cube twice, there are 36 possible outcomes (6 possible outcomes on the first roll, and 6 possible outcomes on the second roll).
We want to find the probability of two specific events happening in this order: first rolling an even number, and then not rolling a 2.
There are three even numbers on a number cube: 2, 4, and 6. So the probability of rolling an even number on the first roll is 3 out of 6, or 1/2.
There are five numbers that are not 2: 1, 3, 4, 5, and 6. So the probability of not rolling a 2 on the second roll is 5 out of 6.
To find the probability of both events happening, we multiply the probabilities:
P(even, then not 2) = (1/2) x (5/6) = 5/12
So the probability of rolling an even number on the first roll and not rolling a 2 on the second roll is 5/12.
4 answers