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# A certain coin with heads on one side and tails on the other has a 1/2

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Math Expert
Joined: 02 Sep 2009
Posts: 58427
A certain coin with heads on one side and tails on the other has a 1/2  [#permalink]

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17 Dec 2018, 01:00
00:00

Difficulty:

25% (medium)

Question Stats:

75% (01:19) correct 25% (01:45) wrong based on 52 sessions

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A certain coin with heads on one side and tails on the other has a $$\frac{1}{2}$$ probability of landing on heads. If the coin is flipped 5 times, how many distinct outcomes are possible if the last flip must be heads? Outcomes are distinct if they do not contain exactly the same results in exactly the same order.

A. 8
B. 12
C. 14
D. 16
E. 32

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Joined: 27 Dec 2016
Posts: 309
Re: A certain coin with heads on one side and tails on the other has a 1/2  [#permalink]

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17 Dec 2018, 19:27
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Joined: 18 Aug 2017
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Location: India
Concentration: Sustainability, Marketing
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Re: A certain coin with heads on one side and tails on the other has a 1/2  [#permalink]

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18 Dec 2018, 01:58
1
Bunuel wrote:
A certain coin with heads on one side and tails on the other has a $$\frac{1}{2}$$ probability of landing on heads. If the coin is flipped 5 times, how many distinct outcomes are possible if the last flip must be heads? Outcomes are distinct if they do not contain exactly the same results in exactly the same order.

A. 8
B. 12
C. 14
D. 16
E. 32

chetan2u :
Hi could you please review this question and suggest on why e is wrong over d?
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Re: A certain coin with heads on one side and tails on the other has a 1/2  [#permalink]

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18 Dec 2018, 06:37
3
Given that the last coin flip is a definite outcome, we should ignore this distraction and focus on the first 4 flips. Since the question states that same outcomes (i.e. number of H or T) in different orders mean distinct outcomes, ordering is important and we need to account for repeats.

For the first 4 flips, the following scenarios are possible:
1) H H H H --> 4!/4! = 1 way of arranging the outcomes.
2) H H H T --> 4!/3! = 4 ways
3) H H T T --> 4!/(2!2!) = 6 ways
4) H T T T --> 4!/(3!) = 4 ways
5) T T T T --> 4!/4! = 1 way
Sum of distinct outcomes = 1 + 4 + 6 + 4 + 1 = 16 ways (Answer: D)
Re: A certain coin with heads on one side and tails on the other has a 1/2   [#permalink] 18 Dec 2018, 06:37
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