To solve this problem, we can use the normal approximation to the binomial distribution since we have a large number of trials.
Let X be the number of fours rolled in 6000 trials. X follows a binomial distribution with n = 6000 trials and p = 1/6 probability of rolling a four in each trial.
The mean of the binomial distribution is np = 6000*(1/6) = 1000 and the standard deviation is sqrt(np(1-p)) = sqrt(1000*(5/6)) ≈ 15.81
To find the probability of getting at least 1050 fours, we can convert this into a z-score using the above mean and standard deviation.
Z = (1050 - 1000) / 15.81 ≈ 3.16
Looking up the z-score in a standard normal distribution table, we find that the probability of Z being greater than 3.16 is approximately 0.001.
So, the approximate probability of getting at least 1050 fours in 6000 rolls of the die is 0.001 or 0.1%.
Therefore, none of the given options are correct.
A fair die is rolled 6,000 times. Assume the trials are independent and the probability of 2ss = 1/6 * of each trial. What's the approximate probability of getting at least 1050 fours in your 6,000 rolls of the die?
A. 044
Β. .048
C. 035
D..05
Ε. 063
1 answer