A man rolls an ordinary die 6 times. What is the probability that he gets at least 5 threes? Round your answer to four decimal places.
in one town, 37% of all voters are democrats. if two voters are randomly selected for a survey, find the probability that they are both democrats.
I got 0.0001500
To find the probability of getting at least 5 threes on rolling an ordinary die 6 times, we need to find the probability of getting exactly 5 threes, the probability of getting exactly 6 threes, and add them together.
Let's break it down step by step:
Step 1: Find the probability of getting exactly 5 threes.
To calculate this, we need to consider that there are 6 rolls and we want exactly 5 threes. Since the probability of getting a three on a single roll is 1/6, we can use the binomial probability formula:
P(X = k) = (n choose k) * p^k * (1-p)^(n-k)
In this case, n (the number of trials) is 6, k (the number of successful outcomes) is 5, and p (the probability of success on a single trial) is 1/6.
The formula becomes:
P(X = 5) = (6 choose 5) * (1/6)^5 * (5/6)^(6-5)
Step 2: Find the probability of getting exactly 6 threes.
Using the same formula as above, we can calculate the probability of getting exactly 6 threes:
P(X = 6) = (6 choose 6) * (1/6)^6 * (5/6)^(6-6)
Step 3: Add the probabilities of getting exactly 5 threes and exactly 6 threes, obtained from steps 1 and 2 respectively:
P(at least 5 threes) = P(X = 5) + P(X = 6)
Step 4: Round the result to four decimal places.
Now, let's calculate it:
Step 1: P(X = 5) = (6 choose 5) * (1/6)^5 * (5/6)^(6-5)
= (6 choose 5) * (1/6)^5 * (5/6)^1
= 6 * (1/7776) * (5/6)
= 5/2592
Step 2: P(X = 6) = (6 choose 6) * (1/6)^6 * (5/6)^(6-6)
= 1 * (1/46656) * (1)
= 1/46656
Step 3: P(at least 5 threes) = P(X = 5) + P(X = 6)
= 5/2592 + 1/46656
= 47/46656
Step 4: Rounding to four decimal places, the final probability becomes approximately 0.0010.
Therefore, the probability that the man gets at least 5 threes when rolling an ordinary die 6 times is approximately 0.0010.