SOLUTION: 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 pr

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

RELATED QUESTIONS
Problem VI - [14 points] A “fair,” “relatively thick” coin, when flipped, can... (answered by Edwin McCravy)

When a coin is fair it can be shown that the number of heads h that appears when you flip (answered by richard1234)

Can You Please Help Me With This Probability Question: A fair coin is flipped once and (answered by Alan3354)

Show all the possible outcomes of tossing two coins Heads-H Tails T Event 1... (answered by stanbon)

A coin is flipped 17 times. What is probability the number of heads will equal... (answered by sofiyac)

A coin is flipped 10 times in a row. a. How many sequences of heads and tails are in... (answered by psbhowmick)

6.8 #17 Can someone please walk me through this? Before a football game, a coin... (answered by Alan3354,Theo)

A coin is weighted so that there is a 62% chance that it will come up "heads" when... (answered by ikleyn)

a fair coin is flipped 3 times and a sequence of heads and tails is recorded. What is the (answered by Susan-math)