a fair die is rolled 3 times. A 4 is considered "success", while all other outcomes are "failures." find the probability of the number of success
a) 2
b)3
Use binomial distribution, which requires a known and constant probability of success, a known number of trials, independence of trials.
probability of success, p=1/6
number of trials, n=3
the probability of x successes is
P(x)=C(n,x)p^x (1-p)^(n-x)
where
C(n,x) is number of combinations choosing x objects out of n, and
C(n,x)=n!/(x!(n-x)!)
(a) x=2 successes
P(2)=C(3,2)(1/6)^2 (5/6)^1
=(6/(2*1))*(1/6)^2*(5/6)
=5/72
(b) x=3 successes
P(3)=C(3,3)(1/6)^3(5/6)^0
=(1)(1/216)(1)
=1/216
Well, well, well, let's talk about probabilities and dice rolls!
To find the probability of a certain number of successes, we need to use a little bit of math magic.
a) Two successes:
Out of the six faces of a die, only one of them is a "4". So, if we want exactly two successes, we need to choose two out of the three rolls to be a "4", and the remaining one to be a failure.
The probability of rolling a "4" is 1/6, and the probability of rolling anything other than a "4" is 5/6. Since we want two "4"s and one failure, we get:
P(2 successes) = (1/6) * (1/6) * (5/6) = 5/216
b) Three successes:
To have three successes, we need all three rolls to be "4". So, the probability of rolling a "4" three times in a row is:
P(3 successes) = (1/6) * (1/6) * (1/6) = 1/216
There you have it! The probability of two successes is 5/216, and the probability of three successes is 1/216.
To find the probability of the number of successes when a fair die is rolled 3 times, we can use the binomial probability formula.
The binomial probability formula is:
P(X = k) = (nCk) * p^k * (1-p)^(n-k)
Where:
- P(X = k) represents the probability of exactly k successes.
- n is the number of trials.
- k is the number of successes.
- p is the probability of success.
In this case:
- n = 3 (the die is rolled 3 times)
- p = 1/6 (since there is 1 success (rolling a 4) out of 6 possible outcomes)
a) To find the probability of getting exactly 2 successes:
P(X = 2) = (3C2) * (1/6)^2 * (5/6)^(3-2)
= 3 * (1/36) * (5/6)
= 15/216
= 5/72
b) To find the probability of getting exactly 3 successes:
P(X = 3) = (3C3) * (1/6)^3 * (5/6)^(3-3)
= 1 * (1/216) * 1
= 1/216
Therefore:
a) The probability of getting exactly 2 successes is 5/72.
b) The probability of getting exactly 3 successes is 1/216.
To find the probability of a specific number of successes when rolling a fair die 3 times, we can use the concept of probability and counting.
Let's break down each part of the problem step by step.
a) To find the probability of exactly 2 successes (rolling a 4), we need to consider all the possible outcomes.
First, let's determine the total number of possible outcomes when rolling a fair die 3 times. Since each roll has 6 possible outcomes (numbers 1 to 6), there are a total of 6 × 6 × 6 = 216 possible outcomes.
To calculate the number of successful outcomes (rolling a 4) when rolling a fair die, we have to consider that each roll is an independent event. So, for each roll, there is a 1 in 6 chance of rolling a 4.
To get exactly 2 successes, we need to find the number of ways we can get two 4s and one non-4 outcome.
To calculate this, we use the concept of combinations, which is denoted as C(n, r). The formula for calculating combinations is:
C(n, r) = n! / (r!(n - r)!)
Here, n represents the total number of items (6 in our case), and r represents the number of items we want to choose (2 in our case).
So, C(6, 2) = 6! / (2!(6 - 2)!) = (6 × 5 × 4 × 3 × 2 × 1) / ((2 × 1) × (4 × 3 × 2 × 1)) = 15
Therefore, the number of ways we can get exactly 2 successes is 15.
Now that we know the number of successful outcomes and the total number of outcomes, we can calculate the probability:
P(2 successes) = Number of successful outcomes / Total number of outcomes = 15 / 216 = 5/72
So, the probability of getting exactly 2 successes (rolling a 4) when rolling a fair die 3 times is 5/72.
b) To find the probability of exactly 3 successes (rolling three 4s), we repeat the same process.
The number of ways we can get exactly 3 successes is given by C(6, 3) = 6! / (3!(6 - 3)!) = (6 × 5 × 4 × 3 × 2 × 1) / ((3 × 2 × 1) × (3 × 2 × 1)) = 20
Therefore, the number of ways we can get exactly 3 successes is 20.
Now we can calculate the probability:
P(3 successes) = Number of successful outcomes / Total number of outcomes = 20 / 216 = 5/54
So, the probability of getting exactly 3 successes (rolling three 4s) when rolling a fair die 3 times is 5/54.