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05 Jul 2024, 01:30
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A
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Rosalie tosses a coin a number of times and counts the total number of times the coin lands tail-side up. How many times does she toss the coin?
(1) The chance that the coin lands tail-side up exactly once is between 0.1 and 0.2.
(2) The chance that the coin lands head side up exactly once is the same as the chance that the coin lands tail-side up exactly once.
(1) The chance that the coin lands tail-side up exactly once is between 0.1 and 0.2.
(2) The chance that the coin lands head side up exactly once is the same as the chance that the coin lands tail-side up exactly once.
05 Jul 2024, 04:45
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(1) The chance that the coin lands tail-side up exactly once is between 0.1 and 0.2.
The chances of landing one tails and an infinite number of heads will be \(\frac{1}{2}*(\frac{1}{2})^{n-1}*\frac{n!}{(n-1)!}\)
The \(\frac{1}{2}\) establishes that one tails is tossed.
\((\frac{1}{2})^{n-1}\) will be for the number of heads, where \(n\) is the number of times the coin is flipped.
Finally, \(\frac{n!}{(n-1)!}\) will ensure all possibilities, without which one will find the probability of getting a tails first and then only heads.
Simplifying \(\frac{1}{2}*(\frac{1}{2})^{n-1}*\frac{n!}{(n-1)!}\):
\(\frac{n}{2^n}\)
We are told that the chances are between 0.1 and 0.2: \(\frac{1}{10}<\frac{n}{2^n}<\frac{2}{10}\)
Plugging in values for n:
1: \(\frac{1}{2}\) which is the same as \(\frac{5}{10}\) out of range
2: \(\frac{2}{4}\) which is the same as \(\frac{5}{10}\) out of range
3: \(\frac{3}{8}\) which is closer to \(\frac{1}{2}\) than to \(\frac{2}{10}\) out of range
4: \(\frac{4}{16}\) which is the same as \(\frac{2.5}{10}\) out of range
5: \(\frac{5}{32}\) when one changes in the inequality to be \(\frac{3.2}{32}<\frac{n}{2^n}<\frac{6.4}{32}\) one sees that it is in range
6: \(\frac{6}{64}\) which is going to be smaller than \(\frac{1}{10}\) out of range
The coin has to have been flipped 5 times.
SUFFICIENT
(2) The chance that the coin lands head side up exactly once is the same as the chance that the coin lands tail-side up exactly once.
As the chances are the same, it is impossible that one make two equations equal to one another when they will cancel out.
INSUFFICIENT
ANSWER A
The chances of landing one tails and an infinite number of heads will be \(\frac{1}{2}*(\frac{1}{2})^{n-1}*\frac{n!}{(n-1)!}\)
The \(\frac{1}{2}\) establishes that one tails is tossed.
\((\frac{1}{2})^{n-1}\) will be for the number of heads, where \(n\) is the number of times the coin is flipped.
Finally, \(\frac{n!}{(n-1)!}\) will ensure all possibilities, without which one will find the probability of getting a tails first and then only heads.
Simplifying \(\frac{1}{2}*(\frac{1}{2})^{n-1}*\frac{n!}{(n-1)!}\):
\(\frac{n}{2^n}\)
We are told that the chances are between 0.1 and 0.2: \(\frac{1}{10}<\frac{n}{2^n}<\frac{2}{10}\)
Plugging in values for n:
1: \(\frac{1}{2}\) which is the same as \(\frac{5}{10}\) out of range
2: \(\frac{2}{4}\) which is the same as \(\frac{5}{10}\) out of range
3: \(\frac{3}{8}\) which is closer to \(\frac{1}{2}\) than to \(\frac{2}{10}\) out of range
4: \(\frac{4}{16}\) which is the same as \(\frac{2.5}{10}\) out of range
5: \(\frac{5}{32}\) when one changes in the inequality to be \(\frac{3.2}{32}<\frac{n}{2^n}<\frac{6.4}{32}\) one sees that it is in range
6: \(\frac{6}{64}\) which is going to be smaller than \(\frac{1}{10}\) out of range
The coin has to have been flipped 5 times.
SUFFICIENT
(2) The chance that the coin lands head side up exactly once is the same as the chance that the coin lands tail-side up exactly once.
As the chances are the same, it is impossible that one make two equations equal to one another when they will cancel out.
INSUFFICIENT
ANSWER A
05 Jul 2024, 05:33
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I think A,
Rosalie tosses a coin a number of times and counts the total number of times the coin lands tail-side up.
How many times does she toss the coin?
1st - The chance that the coin lands tail-side up exactly once is between 0.1 and 0.2.
Probability of showing head/tail when one fair coin tossed = p = 0.5
The probability of getting exactly one tail in n tosses is given by the binomial distribution formula:
P(exactly one tail) = nC1 * p * (1-p)^(n-1) = n*0.5*0.5^n-1 = n*(0.5^n)
This is in between 0.1 and 0.2
0.1 < n*(0.5^n) < 0.2
lets check now putting values for n=1,2,3,4....
n=1, n*(0.5^n) = 0.5
n=2, n*(0.5^n) = 2*0.25 = 0.5
n=3, n*(0.5^n) = 3*0.125 = 0.375
n=4, n*(0.5^n) = 4*0.0625 = 0.25
n=5, n*(0.5^n) = 5*0.03125 = 0.15625 (within given range 0.1 < n*(0.5^n) < 0.2)
n=6, n*(0.5^n) = 6*0.015625 = 0.09375
Hence Sufficient.
2nd - The chance that the coin lands head side up exactly once is the same as the chance that the coin lands tail-side up exactly once.
This statement will lead us to nothing, Since the probability of getting exactly one head and exactly one tail is the same and equals with the same binomial distribution logic = n*(0.5^n) and when we equate both will just cancels each other and we cant calculate value for n. Not sufficient.
Rosalie tosses a coin a number of times and counts the total number of times the coin lands tail-side up.
How many times does she toss the coin?
1st - The chance that the coin lands tail-side up exactly once is between 0.1 and 0.2.
Probability of showing head/tail when one fair coin tossed = p = 0.5
The probability of getting exactly one tail in n tosses is given by the binomial distribution formula:
P(exactly one tail) = nC1 * p * (1-p)^(n-1) = n*0.5*0.5^n-1 = n*(0.5^n)
This is in between 0.1 and 0.2
0.1 < n*(0.5^n) < 0.2
lets check now putting values for n=1,2,3,4....
n=1, n*(0.5^n) = 0.5
n=2, n*(0.5^n) = 2*0.25 = 0.5
n=3, n*(0.5^n) = 3*0.125 = 0.375
n=4, n*(0.5^n) = 4*0.0625 = 0.25
n=5, n*(0.5^n) = 5*0.03125 = 0.15625 (within given range 0.1 < n*(0.5^n) < 0.2)
n=6, n*(0.5^n) = 6*0.015625 = 0.09375
Hence Sufficient.
2nd - The chance that the coin lands head side up exactly once is the same as the chance that the coin lands tail-side up exactly once.
This statement will lead us to nothing, Since the probability of getting exactly one head and exactly one tail is the same and equals with the same binomial distribution logic = n*(0.5^n) and when we equate both will just cancels each other and we cant calculate value for n. Not sufficient.
