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27 May 2018, 10:19
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
71% (01:20) correct
29%
(01:29)
wrong
based on 466
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How many times was a fair coin tossed?
(1) If the coin has been tossed 4 times fewer, the probability of getting heads on every toss would have been (1/8)
(2) When a coin is tossed the number of times, the number of different possible sequences of heads and tails is 128.
(1) If the coin has been tossed 4 times fewer, the probability of getting heads on every toss would have been (1/8)
(2) When a coin is tossed the number of times, the number of different possible sequences of heads and tails is 128.
28 May 2018, 05:48
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Bunuel
1. When a fair coin is tossed once, the probability of getting heads on that toss is \(\frac{1}{2}\)
Similarly, when the fair coin is twice, the probability of getting heads on that toss is \(\frac{1}{2}*\frac{1}{2}\) = \(\frac{1}{4}\)
So, with every iteration of tossing the fair coin, we have an increase of \(\frac{1}{2}\)
If the coin was tossed 4 times fewer and the probability is \(\frac{1}{8}\)
If x is the number of times that the fair coin was tossed \(\frac{1}{2^{x-4}}\) = \(\frac{1}{2^3}\) -> \(x-4 = 3\) -> \(x = 7\) (Sufficient)
2. Every time a coin is tossed, there are 2 ouitcomes. So, if there are 128 outcomes possible,
and x is the number of times the coin is tossed, we have \(2^x = 128 = 2^7\) -> \(x = 7\) (Sufficient - Option D)
General Discussion
28 May 2018, 07:09
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Solution
To find:
- • How many times the fair coin is tossed
Analysing Statement 1
- • As per the information provided in Statement 1, the probability of getting heads on every toss would have been 1/8 if the coin has been tossed 4 times fewer
- • 1 time = \(\frac{1}{2}\)
• 2 times = \((\frac{1}{2})^2\)
• 3 times = \((\frac{1}{2})^3\)
• Similarly, n times = \((\frac{1}{2})^n\)
- • \((\frac{1}{2})^{n-4} = \frac{1}{8} = (\frac{1}{2})^3\)
Hence, n – 4 = 3
Or, n = 7
Analysing Statement 2
- • As per the information provided in Statement 2, when the coin is tossed the number of times, the number of different possible sequences of heads and tails is 128
- • 1 time = \(2^1\)
• 2 times = \(2^2\)
• 3 times = \(3^2\)
• Similarly, n times = \(2^n\)
- • \(2^n = 128 = 2^7\)
Or, n = 7
Hence, the correct answer is option D.
Answer: D
