In a version of the game of roulette, a steel ball is rolled onto a wheel that contains 19 red, 19 black and 2 green slots. If the ball is rolled 25 times, the probability that it falls into the green slots two or more times:
a) is 5%.
b) is 35.8%.
c) is 12.7%.
d) cannot be determined since there are three possible outcomes of rolling the steel ball.
THE ANSWER IS B.
Can someone explain why.
Thanks
To find the probability that the ball falls into the green slots two or more times out of 25 rolls, we need to use the binomial probability formula.
The binomial probability formula is given by:
P(x successes out of n trials) = C(n, x) * p^x * (1-p)^(n-x)
Where P is the probability, n is the number of trials, x is the number of successes, C(n, x) is the number of combinations of n things taken x at a time, p is the probability of success in a single trial, and (1-p) is the probability of failure in a single trial.
In this case, the number of trials (n) is 25, and we want the probability of getting two or more successes, which means x = 2, 3, 4, ..., 25.
The probability of the ball falling into the green slots in a single trial is given by the ratio of the number of green slots (2) to the total number of slots (19 + 19 + 2 = 40), which is 2/40 = 1/20.
Now, let's calculate the probability using the formula. We need to calculate the probability of getting exactly 2, 3, 4, ..., or 25 successes and then sum them up.
P(2 or more successes) = P(x=2) + P(x=3) + P(x=4) + ... + P(x=25)
P(2 or more successes) = [C(25, 2) * (1/20)^2 * (19/20)^(25-2)] + [C(25, 3) * (1/20)^3 * (19/20)^(25-3)] + [C(25, 4) * (1/20)^4 * (19/20)^(25-4)] + ... + [C(25, 25) * (1/20)^25 * (19/20)^(25-25)]
Using a binomial calculator or statistical software, we can calculate this probability, which is approximately 35.8%.
Therefore, the correct answer is (b) 35.8%.