a fair die is tossed 180 times. Find the probability P that the face 6 will appear between 29 and 32 times inclusive.
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To solve this problem, we need to use the concept of probability and the properties of a fair die. Let's break it down step by step:
Step 1: Determine the probability of getting a face 6 on a single toss of the die.
Since the die is fair, it has six equally likely outcomes (faces). Out of these six outcomes, only one corresponds to getting a face 6. Therefore, the probability of getting a face 6 on a single toss is 1/6.
Step 2: Define the number of trials and the desired range of occurrences.
The problem states that the die is tossed 180 times and we want to find the probability of getting a face 6 between 29 and 32 times (inclusive). This means the number of occurrences falls within this range.
Step 3: Calculate the probability of getting a specific number of occurrences.
To find the probability of exactly k occurrences of face 6 out of n trials, we need to use the binomial probability formula:
P(k) = (n choose k) * p^k * (1-p)^(n-k)
In our case, n = 180 (number of tosses), k can be 29, 30, 31, or 32 (desired range), and p = 1/6 (probability of getting face 6 on a single toss).
Step 4: Calculate the final probability.
To find the probability within the desired range (inclusive), we need to sum up the probabilities for the individual cases (k = 29, 30, 31, 32):
P = P(29) + P(30) + P(31) + P(32)
Now, let's calculate:
P(29) = (180 choose 29) * (1/6)^29 * (5/6)^(180-29)
P(30) = (180 choose 30) * (1/6)^30 * (5/6)^(180-30)
P(31) = (180 choose 31) * (1/6)^31 * (5/6)^(180-31)
P(32) = (180 choose 32) * (1/6)^32 * (5/6)^(180-32)
Finally, calculate the sum P = P(29) + P(30) + P(31) + P(32).