In a particular roulette game, there's a 1/36 chance of winning. in a single day, a gambler plays 100 rounds, and wins in 7 of them. what's the p-value for winning 7 or more out of 100 rounds? (hint: be sure you know which kind of probability distribution you're dealing with before you do the calculation.)

0.021739

Step-by-step explanation:

This situation can be modeled with the Binomial Distribution which gives the probability of an event that occurs exactly k times out of n, and is given by

where  

= combination of n elements taken k at a time.

p = probability that the event (“success”) occurs once

q = 1-p

In this case, the event “success” is winning the roulette game with probability 1/36 = 0.027777 and n=100 rounds.

The probability value of winning the roulette game 7 or more times out of 100, is  

P(7;100)+P(8;100)+...+P(100;100) =

1 - P(0;100)+P(1;100)+...+P(6;100)

We can find the sum of these last 7 terms either by hand or computer-assisted and we would find  

P(0;100)+P(1;100)+...+P(6;100) =

and  the p-value (probability value) for winning 7 or more out of 100 rounds is

1-0.978261 = 0.021739

option: d is the correct answer.

d) none of the above.

step-by-step explanation:

clearly from the figure attached to the question we could see that there is no shape given. hence without any shape given in the figure and after rotation we can't obtained a square pyramid.

since, to obtain a image we need a proper figure of the pre-image so that it can be transformed.

hence, the correct answer is:

d) none of the above.

step-by-step explanation: