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20 Aug 2018, 05:50
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B
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38% (02:04) correct
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Minimum of how many people are needed to have the probability of more than 1/2 that at lease one of them was born on either on Monday or on Tuesday?
A. 2
B. 3
C. 4
D. 5
E. 6
M37-16
A. 2
B. 3
C. 4
D. 5
E. 6
M37-16
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19 Nov 2021, 04:22
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Official Solution:
Minimum of how many people are needed to have the probability of more than 1/2 that at lease one of them was born on either on Monday or on Tuesday?
A. \(2\)
B. \(3\)
C. \(4\)
D. \(5\)
E. \(6\)
The probability that a person is born on either on Monday or on Tuesday is \(\frac{2}{7}\) and the probability that a person is NOT born on either on Monday or on Tuesday is therefore \(\frac{5}{7}\).
Say minimum \(n\) people are needed to have the probability of more than 1/2 that at lease one of them was born on either on Monday or on Tuesday. In this case the probability that NONE of them is born on either on Monday or on Tuesday will be \((\frac{5}{7})^n\). Thus, the probability that AT LEAST ONE of them is born on either on Monday or on Tuesday will be \(1-(\frac{5}{7})^n\).
The question asks to find minimum \(n\) such that \(1-(\frac{5}{7})^n > \frac{1}{2}\):
\( \frac{1}{2}>(\frac{5}{7})^n\);
The least value of \(n\) that satisfies this is 3: \((\frac{5}{7})^3=\frac{125}{343} < \frac{1}{2}\).
Answer: B
Minimum of how many people are needed to have the probability of more than 1/2 that at lease one of them was born on either on Monday or on Tuesday?
A. \(2\)
B. \(3\)
C. \(4\)
D. \(5\)
E. \(6\)
The probability that a person is born on either on Monday or on Tuesday is \(\frac{2}{7}\) and the probability that a person is NOT born on either on Monday or on Tuesday is therefore \(\frac{5}{7}\).
Say minimum \(n\) people are needed to have the probability of more than 1/2 that at lease one of them was born on either on Monday or on Tuesday. In this case the probability that NONE of them is born on either on Monday or on Tuesday will be \((\frac{5}{7})^n\). Thus, the probability that AT LEAST ONE of them is born on either on Monday or on Tuesday will be \(1-(\frac{5}{7})^n\).
The question asks to find minimum \(n\) such that \(1-(\frac{5}{7})^n > \frac{1}{2}\):
\( \frac{1}{2}>(\frac{5}{7})^n\);
The least value of \(n\) that satisfies this is 3: \((\frac{5}{7})^3=\frac{125}{343} < \frac{1}{2}\).
Answer: B
02 Oct 2018, 22:37
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Bunuel
GMAT CLUB TESTS' FRESH QUESTION:
Minimum of how many people are needed to have the probability of more than 1/2 that at lease one of them was born on either on Monday or on Tuesday?
A. 2
B. 3
C. 4
D. 5
E. 6
As, question stated that at least one of the people have to have birthday on either Monday or Tuesday, I need to figure out the probability of all of them having birthday on remaining 5 days of week (other than Monday and Tuesday), then simply asses the value whether it is less than 0.5. If it is less than 0.5 then we have our answer.
First option 2 people: probability of both of them having birthday of remaining 5 days out of 7 days : \(\frac{5}{7} * \frac{5}{7}\) =\(\frac{25}{49}\) > 0.5
2nd Option 3 people: probability of all 3 of them having birthday of remaining 5 days out of 7 days : \(\frac{5}{7}*\frac{5}{7}*\frac{5}{7} = \frac{125}{363}\)< 0.5 ---> so 3 people is our answer or option [B]
