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31 Aug 2024, 03:56
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If an unfair coin is flipped 16 times, the probability of getting at least one head is \(\frac{15}{16}\). What is the probability of getting at least one head if the coin is flipped 8 times?
A. \(\frac{1}{16}\)
B. \(\frac{1}{8}\)
C. \(\frac{1}{4}\)
D. \(\frac{1}{2}\)
E. \(\frac{3}{4}\)
A. \(\frac{1}{16}\)
B. \(\frac{1}{8}\)
C. \(\frac{1}{4}\)
D. \(\frac{1}{2}\)
E. \(\frac{3}{4}\)
31 Aug 2024, 03:56
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Official Solution:
If an unfair coin is flipped 16 times, the probability of getting at least one head is \(\frac{15}{16}\). What is the probability of getting at least one head if the coin is flipped 8 times?
A. \(\frac{1}{16}\)
B. \(\frac{1}{8}\)
C. \(\frac{1}{4}\)
D. \(\frac{1}{2}\)
E. \(\frac{3}{4}\)
A fair coin has the probability of heads equal to the probability of tails, so \(P(tail) = P(head) = \frac{1}{2}\). We are told that the coin is unfair, so \(P(tail) ≠ P(head)\).
Let's denote the probability of getting tails as \(x\) and the probability of heads as \(1−x\). In this case, the probability of getting at least one head in 16 flips would be \(P(heads \geq 1) = 1 - P(tails = 16) = 1 - x^{16}\). Given that this equals \(\frac{15}{16}\), we have:
\(1 - x^{16} = \frac{15}{16}\)
\(x^{16} = \frac{1}{16}\)
The question asks to find the probability of getting at least one head if the coin is flipped 8 times, which would equal \(P(heads \geq 1) = 1 - P(tails = 8) = 1 - x^{8}\). Since \(x^{16} = \frac{1}{16}\), taking the square root would give \(x^{8} = \frac{1}{4}\). Therefore, \(1 - x^{8} = 1 - \frac{1}{4} = \frac{3}{4}\)
Answer: E
If an unfair coin is flipped 16 times, the probability of getting at least one head is \(\frac{15}{16}\). What is the probability of getting at least one head if the coin is flipped 8 times?
A. \(\frac{1}{16}\)
B. \(\frac{1}{8}\)
C. \(\frac{1}{4}\)
D. \(\frac{1}{2}\)
E. \(\frac{3}{4}\)
A fair coin has the probability of heads equal to the probability of tails, so \(P(tail) = P(head) = \frac{1}{2}\). We are told that the coin is unfair, so \(P(tail) ≠ P(head)\).
Let's denote the probability of getting tails as \(x\) and the probability of heads as \(1−x\). In this case, the probability of getting at least one head in 16 flips would be \(P(heads \geq 1) = 1 - P(tails = 16) = 1 - x^{16}\). Given that this equals \(\frac{15}{16}\), we have:
\(1 - x^{16} = \frac{15}{16}\)
\(x^{16} = \frac{1}{16}\)
Answer: E
04 Nov 2024, 04:16
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Bunuel
Can you please explain how is the square root of X^16 equals X^8?
I get it when you say the suare root of 1/16 is 1/4 BUT NOT how the square root of X^16 equals X^8?
Thanks in advance!
Can you please explain how is the square root of X^16 equals X^8?
I get it when you say the suare root of 1/16 is 1/4 BUT NOT how the square root of X^16 equals X^8?
Thanks in advance!
