Customer arrivals at a fast-food outlet conform to a Poisson distribution with an average rate of 16 customers per hour. In statistical probability analysis, the Poisson distribution is a commonly utilized discrete probability distribution. Employing the formula, it has been calculated that 0.0661 represents the probability of there being precisely 12 arrivals in the next hour.
Answer:
There is a 1.5267% chance that he loses AT MOST 3 times.
Step-by-step explanation:
Each game Andy plays has only two outcomes: a win or a loss. The probability of winning doesn’t depend on previous games, leading us to use the binomial probability distribution for this situation.
Binomial probability distribution
This distribution calculates the probability of achieving exactly x successes in n trials with only two possible outcomes for X.

Where
signifies the various combinations of x objects from a set of n elements, calculated using:

And p represents the likelihood of X occurring.
The winning probability for a certain carnival game is 2 out of 5.
This implies a losing chance of (5-2) in 5, equaling 3 out of 5.
Thus, 
Considering 12 games:
Consequently,
.
What probability is there that he loses AT MOST 3 times?
.
Where:






The likelihood of losing AT MOST 3 times is 1.5267%.
C because a negative would turn into a positive.
Response:
Count of bottles = 6
Count of cans = c
Detailed reasoning:
Based on the following information:
Containers for selling water:
17 oz bottle
11 oz can
Let the number of bottles be denoted as b
Count of cans represented as c
b + c = 11 - - - (1)
17b + 11c = 157 - - - (2)
b = 11 - c
Insert into (2)
17(11 - c) + 11c = 157
187 - 17c + 11c = 157
187 - 6c = 157
-6c = 157 - 187
-6c = -30
c = 30/6
c = 5
From; b = 11 - c
b = 11 - 5
b = 6
Thus,
Count of bottles = 6
Count of cans = c
9×2=18. then append the two zeros: 9×2=18+00=1,800