Electronics and Communication Engineering (ECE) Exam > Electronics and Communication Engineering (ECE) Questions > Eight coins are tossed simultaneously. The pr...
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Eight coins are tossed simultaneously. The probability of getting at least 6 heads is
- a)7/64
- b)37/256
- c)57/64
- d)249/256
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
Eight coins are tossed simultaneously. The probability of getting at l...
p = 1/2, q = 1/2, n = 8. Required probability = P (6 heads or 7 heads or 8 heads)
Most Upvoted Answer
Eight coins are tossed simultaneously. The probability of getting at l...
Understanding the Problem
When tossing eight coins, we want to calculate the probability of getting at least 6 heads. This includes the cases of getting 6, 7, or 8 heads.
Using the Binomial Probability Formula
The probability of obtaining exactly k heads in n tosses is given by:
- P(X = k) = C(n, k) * (p^k) * ((1-p)^(n-k))
Where:
- C(n, k) = n! / (k!(n-k)!)
- n = total number of coins (8 in this case)
- p = probability of getting heads (0.5 for a fair coin)
Calculating the Required Probabilities
We need to calculate the probabilities for k = 6, 7, and 8.
1. For 6 heads:
- C(8, 6) = 28
- P(X = 6) = 28 * (0.5^6) * (0.5^2) = 28 * (1/64) * (1/4) = 28/256
2. For 7 heads:
- C(8, 7) = 8
- P(X = 7) = 8 * (0.5^7) * (0.5^1) = 8 * (1/128) * (1/2) = 8/256
3. For 8 heads:
- C(8, 8) = 1
- P(X = 8) = 1 * (0.5^8) = 1/256
Total Probability of At Least 6 Heads
Now, we sum up the probabilities calculated above:
- P(X ≥ 6) = P(X = 6) + P(X = 7) + P(X = 8)
- P(X ≥ 6) = (28/256) + (8/256) + (1/256) = 37/256
Conclusion
Therefore, the probability of getting at least 6 heads when tossing 8 coins is 37/256, which corresponds to option 'B'.
When tossing eight coins, we want to calculate the probability of getting at least 6 heads. This includes the cases of getting 6, 7, or 8 heads.
Using the Binomial Probability Formula
The probability of obtaining exactly k heads in n tosses is given by:
- P(X = k) = C(n, k) * (p^k) * ((1-p)^(n-k))
Where:
- C(n, k) = n! / (k!(n-k)!)
- n = total number of coins (8 in this case)
- p = probability of getting heads (0.5 for a fair coin)
Calculating the Required Probabilities
We need to calculate the probabilities for k = 6, 7, and 8.
1. For 6 heads:
- C(8, 6) = 28
- P(X = 6) = 28 * (0.5^6) * (0.5^2) = 28 * (1/64) * (1/4) = 28/256
2. For 7 heads:
- C(8, 7) = 8
- P(X = 7) = 8 * (0.5^7) * (0.5^1) = 8 * (1/128) * (1/2) = 8/256
3. For 8 heads:
- C(8, 8) = 1
- P(X = 8) = 1 * (0.5^8) = 1/256
Total Probability of At Least 6 Heads
Now, we sum up the probabilities calculated above:
- P(X ≥ 6) = P(X = 6) + P(X = 7) + P(X = 8)
- P(X ≥ 6) = (28/256) + (8/256) + (1/256) = 37/256
Conclusion
Therefore, the probability of getting at least 6 heads when tossing 8 coins is 37/256, which corresponds to option 'B'.
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Eight coins are tossed simultaneously. The probability of getting at least 6 heads isa)7/64b)37/256c)57/64d)249/256Correct answer is option 'B'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2025 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about Eight coins are tossed simultaneously. The probability of getting at least 6 heads isa)7/64b)37/256c)57/64d)249/256Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Eight coins are tossed simultaneously. The probability of getting at least 6 heads isa)7/64b)37/256c)57/64d)249/256Correct answer is option 'B'. Can you explain this answer?.
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