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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