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
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.