a fair of die is tossed 180 times. Find the probability P that the face 6 will appear between 29 and 32 times inclusive?
To find the probability of getting the face 6 between 29 and 32 times (inclusive) when a fair die is tossed 180 times, we need to consider the binomial distribution.
The binomial distribution formula is given by:
P(X = k) = C(n, k) * p^k * (1-p)^(n-k)
where:
- P(X = k) is the probability of getting exactly k successes
- n is the number of trials (in this case, the number of times the die is tossed)
- k is the number of successes (in this case, the number of times the face 6 appears)
- p is the probability of success in a single trial (in this case, the probability of getting a 6 on a fair die)
- C(n, k) is the number of combinations of n items taken k at a time, which can be calculated as C(n, k) = n!/((n-k)! * k!)
In this case, n = 180, p = 1/6 (since there are six possible outcomes on a fair die, and only one of them is a 6), and we need to calculate P for k = 29, 30, 31, and 32.
Let's calculate P for each value of k and then sum them up to get the final probability.
P(X = 29) = C(180, 29) * (1/6)^29 * (5/6)^(180-29)
P(X = 30) = C(180, 30) * (1/6)^30 * (5/6)^(180-30)
P(X = 31) = C(180, 31) * (1/6)^31 * (5/6)^(180-31)
P(X = 32) = C(180, 32) * (1/6)^32 * (5/6)^(180-32)
Finally, we can sum up these individual probabilities to get the total probability P:
P = P(X = 29) + P(X = 30) + P(X = 31) + P(X = 32)
You can use a calculator or a programming language to calculate the binomial coefficients C(n, k), exponentiations, and summation to get the final answer for P.