a die is rolled 7 times. find the probability of getting at least one 3.
Ah, the probability of rolling a die! It's like a game of chance mixed with an intense desire for that lucky number. Let's unpack your question, shall we?
To find the probability of getting at least one 3 when rolling a die seven times, it's easier to calculate the probability of not getting a 3 in any of the rolls and then subtract it from one (since the sum of all probabilities is 1).
The probability of not rolling a 3 on a single roll is 5/6 (since there are six sides on a die and only one of them is a 3). So, the probability of not rolling a 3 in all seven rolls is (5/6)^7.
Now, let's subtract that from one to find the probability of getting at least one 3: 1 - (5/6)^7. And if I had to assign a mathematical value to my clownish gut feeling, I'd guess it's around 0.665 or approximately 66.5%. So, you've got a pretty decent chance of rolling that lucky 3 at least once in seven rolls!
But hey, don't worry if you don't roll a 3 right away. Remember, life is like a die; it always has a way of rolling with the ups and downs!
so we exclude the probability of "no 3"
prob of no 3 in 7 tries = (5/6)^7
prob of your event = 1 - (5/6)^7 = .7209
To find the probability of getting at least one 3 when rolling a die 7 times, we can use the concept of complementary probability.
First, let's determine the probability of not getting a 3 on a single roll.
There are a total of 6 possible outcomes when rolling a die (the numbers 1, 2, 3, 4, 5, or 6), and only one of them is a 3. Therefore, the probability of not getting a 3 on a single roll is:
P(not getting a 3) = (6 - 1) / 6 = 5 / 6
Now, let's consider the probability of not getting a 3 on any of the 7 rolls. Since each roll is independent, we can multiply the probabilities together:
P(not getting a 3 on any of the 7 rolls) = (5 / 6) * (5 / 6) * (5 / 6) * (5 / 6) * (5 / 6) * (5 / 6) * (5 / 6) = (5/6)^7
Finally, to find the probability of getting at least one 3, we subtract the probability of not getting any 3s from 1 (complementary probability):
P(getting at least one 3) = 1 - P(not getting a 3 on any of the 7 rolls) = 1 - (5/6)^7
Now, let's calculate the probability:
P(getting at least one 3) = 1 - (5/6)^7 ≈ 0.665
Therefore, the probability of getting at least one 3 when rolling a die 7 times is approximately 0.665, or 66.5%.
To find the probability of getting at least one 3 when a die is rolled 7 times, we can use the concept of complementary probability.
The probability of something happening is equal to 1 minus the probability of it not happening. In this case, the complement is the event of not getting any 3 in the 7 rolls.
The probability of not getting a 3 on a single roll of the die is 5/6 (since there are 6 possible outcomes and only 1 outcome is a 3). So, the probability of not getting a 3 in 7 rolls is (5/6)^7.
Now, to find the probability of getting at least one 3, we subtract the probability of not getting any 3 from 1:
P(at least one 3) = 1 - P(no 3) = 1 - (5/6)^7
Using a calculator or any programming language, we can evaluate this expression to find the numerical value of the probability.