The probability that a 2 or 3 is rolled at least once is 0.14.
<h3>
Probability of not rolling a 2 or 3 in a single die roll of two dice</h3>
when a six die is rolled once there is 11/12 probability of not rolling 2 or 3.
rolling a six sided die two times, p(2 or 3)' = 11/12
rolling a six sided die four times, p(2 or 3)' = (11/12)² = 121/144
the probability that a 2 or 3 is rolled at least once is calculated as follows;
P = 1 - p(2 or 3)'
P = 1 - 121/144
P = 20/144
P = 0.14
Thus, the probability that a 2 or 3 is rolled at least once is 0.14.
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X=-1/9 I used a calculator
Using the binomial distribution, there is a 0.0874 = 8.74% probability that not enough seats will be available.
<h3>What is the binomial distribution formula?</h3>
The formula is:


The parameters are:
- x is the number of successes.
- n is the number of trials.
- p is the probability of a success on a single trial.
For this problem, the values of the parameters are given by:
n = 15, p = 0.85.
The probability that not enough seats will be available is P(X = 15), as the only outcome in which not enough seats will be available is when all 15 people who bought the ticket show up, hence:


0.0874 = 8.74% probability that not enough seats will be available.
More can be learned about the binomial distribution at brainly.com/question/24863377
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I'm pretty sure the answer is .32
Answer:
Step-by-step explanation:
32 - 7(6+12) / 4^2
25 + 18 /4^2
43 / 16
2.6875 when you round it you get 2.7