Aroller coaster has 3 seats in each of 11 rows. riders are assigned to seats in the order that they arrive. if you ride this roller coaster once, what is the probability of getting the coveted first row? how many times must you ride in order to have at least a 94% chance of getting a first-row seat at least once?

You must ride it 30 times at least

Step-by-step explanation:

Independently of the order, you know that only 3 persons out of 33 get seats in the first row (Even if there werent 33 persons to fill the roller coaster, you can fill the remaining seats with as reserved place). The probability of you being one of those 3 persons that got the first row is 3 out of 33, hence 3/33 = 1/11 = 0.0909.

If you ride the roller coaster n times, each time you will have a 9.0909% chance of getting in the first row. The probability of getting a first row seat at least once can be computed by calculing the complementary event: never getting a seat, and substract from 1 that probability. That probability is the probability of never getting a first row seat on a single ride powered by n times, hence it is (1-0.0909)^n = 0.90909^n.

We want n such that 1-0.90909^n ≥0.94, or quivalently, 0.90909^n = 0.06

Therefore,

[tex] n = \frac{ln(0.06)}{ln(0.90909)} = 29.518

We take n = 30.

picture below

step-by-step explanation:

because 1/3 is a fraction. we can tell because there is a divisor symbol between the 1 and the 3. if we were to add 2 fractions together, we would add the numerators, which is the number on top, but not the denominators, which is the number on the bottom

1/3+1/3=2/3

hope this !