Question 1186719: A “fair,” “relatively thick” coin, when flipped, can show heads (H) or tails (T) with equal probabilities, but it is also probable to come to rest on its edge (E) with a probability of 4.5% (0.045). Showing your work, answer the following questions
1. Set up a probability distribution table for the three possible events.
2. If the coin were flipped twice, what would be the probability that it would show tails the first time and would come to rest on its edge the second time? Provide a brief explanation to support your computation(s). Show the probability value in 4 decimal places.
3. If the coin were flipped 14 times, what would be the probability that the coin would come to rest on its edge (“getting edge”) at most twice? Showing your work, report the probability value in 4 decimal places. Treat the problem as a binomial experiment.
4. Determine the probability of getting heads between 4 to 6 times (both inclusive), i.e., 4 times, 5 times, or 6 times, when the coin is flipped 14 times. Showing your work, report the probability value in 4 decimal places. Treat the problem as a binomial experiment.
5. What is the probability of getting “heads or tails” at least once when the coin is flipped 14 times? Showing your work, report the probability value in 4 decimal places. Treat the problem as a binomial experiment.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's the solution:
**1. Probability Distribution Table:**
| Outcome | Probability |
|---|---|
| Heads (H) | 0.4775 |
| Tails (T) | 0.4775 |
| Edge (E) | 0.045 |
*Explanation:* The probabilities of heads and tails must be equal since the coin is fair. Since the total probability must equal 1, we have: P(H) + P(T) + P(E) = 1. Therefore, 2P(H) + 0.045 = 1, so P(H) = P(T) = (1 - 0.045) / 2 = 0.4775.
**2. Probability of Tails then Edge:**
The flips are independent events. Therefore, we multiply the probabilities:
P(T then E) = P(T) * P(E) = 0.4775 * 0.045 = 0.0214875 ≈ 0.0215
**3. Probability of at most two edges in 14 flips:**
This is a binomial probability problem. Let X be the number of times the coin lands on its edge. We want to find P(X ≤ 2). The probability of "success" (landing on edge) is p = 0.045, and the number of trials is n = 14.
P(X ≤ 2) = P(X=0) + P(X=1) + P(X=2)
The binomial probability formula is: P(X=k) = (nCk) * p^k * (1-p)^(n-k), where nCk is "n choose k".
* P(X=0) = (14C0) * (0.045)^0 * (0.955)^14 ≈ 0.5204
* P(X=1) = (14C1) * (0.045)^1 * (0.955)^13 ≈ 0.3587
* P(X=2) = (14C2) * (0.045)^2 * (0.955)^12 ≈ 0.1090
P(X ≤ 2) ≈ 0.5204 + 0.3587 + 0.1090 ≈ 0.9881
**4. Probability of 4 to 6 heads in 14 flips:**
This is also a binomial probability problem. The probability of "success" (getting heads) is p = 0.4775, and n = 14. We want to find P(4 ≤ X ≤ 6) = P(X=4) + P(X=5) + P(X=6).
* P(X=4) = (14C4) * (0.4775)^4 * (0.5225)^10 ≈ 0.1920
* P(X=5) = (14C5) * (0.4775)^5 * (0.5225)^9 ≈ 0.2253
* P(X=6) = (14C6) * (0.4775)^6 * (0.5225)^8 ≈ 0.1966
P(4 ≤ X ≤ 6) ≈ 0.1920 + 0.2253 + 0.1966 ≈ 0.6139
**5. Probability of at least one "heads or tails" in 14 flips:**
Since the only other outcome is "edge," getting "heads or tails" is the complement of getting "edge" on all 14 flips.
P(at least one H or T) = 1 - P(all edges)
P(all edges) = (0.045)^14 ≈ 2.91 x 10⁻¹⁸
P(at least one H or T) = 1 - (0.045)^14 ≈ 1.0000
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