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If six coins are flipped simultaneously, the probability of

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If six coins are flipped simultaneously, the probability of [#permalink]

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New post 23 Jan 2014, 13:37
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63% (01:04) correct 37% (01:05) wrong based on 131 sessions

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If six coins are flipped simultaneously, the probability of getting at least one heads and at least one tails is closest to

A) 3%
B) 6%
C) 75%
D) 94%
E) 97%
[Reveal] Spoiler: OA

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Re: If six coins are flipped simultaneously, the probability of [#permalink]

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New post 23 Jan 2014, 13:38
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Probability of getting one head (or one tail) = 1/2.

Case one: 1st die is head. Probability of at least one of the remaining five to be tail = 1 - P(none tail).
Probability(none tail i.e. all head) = 1/(2^5) = 1/32. Thus, P(at least one tail) = 31/32.
Probability of case one that one is head and at least another is tail,
i.e. at least one head and one tail = 1/2 * 31/32 = 31/64.

Case two: 1st die is tail. Probability of at least one of the remaining five to be head = 1 - P(none head).
Probability(none head i.e. all tail) = 1/(2^5) = 1/32. Thus, P(at least one head) = 31/32.
Probability of case two that one is tail and at least another is head,
i.e. at least one tail and one head = 1/2 * 31/32 = 31/64. .....we don't need to do this second case calculation
actually since we know it will be the same result as tail and head have same probability.

Since the events that could happen are Case One or Case Two we have to ADD these two probabilities:
31/64 + 31/64 = 62/64.

Two question regarding this:
1. Is there a faster approach than the aforementioned (assuming the rationale is correct).

2. How do I figure out quickly that 62/64 = 31/32 = 0.96875 = 97%?

I was lucky on this practice question as I was running out of time, and my gut feeling said it has to be close to 99%.
But just rounding down the numerator to make division easier ... 30/32 = 0.9375 = 94% answer would change to D.


Correct answer is btw. E.

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Re: If six coins are flipped simultaneously, the probability of [#permalink]

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New post 23 Jan 2014, 13:45
BabySmurf wrote:
Two question regarding this:
1. Is there a faster approach than the aforementioned (assuming the rationale is correct).

2. How do I figure out quickly that 62/64 = 31/32 = 0.96875 = 97%?

I was lucky on this practice question as I was running out of time, and my gut feeling said it has to be close to 99%.
But just rounding down the numerator to make division easier ... 30/32 = 0.9375 = 94% answer would change to D.


Correct answer is btw. E.


1. Yes. You can reword the question to: What is the probability you won't get all heads or all tails?

Should be fairly straightforward that probability of all heads is 1/(2^6). Likewise for all tails. so 1/64 + 1/64 = 2/64... and thus the probability of not getting all heads or all teails is 1 - 1/32 or 31/32.

2. 32 is close to 33... and 33*3 is pretty close to 100. So I would do 31/32 = 93/96. Since we're just four off, 93+4 / 96+4 or 97/100 should be close enough.
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Re: If six coins are flipped simultaneously, the probability of [#permalink]

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New post 23 Jan 2014, 19:13
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Yup, I attempted to use the reverse combinatorics approach (I think that's what it is called) and it was pretty quick.

total potential outcomes (H or T @6 Coins): 2^6 = 64

outcomes that don't include at least one heads and one tails:

\HHHHHH
TTTTTT

so two. then take 1-P(not HT)

1 - 2/64 = 31/32 = 93/96 = or very close to 97%
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Re: If six coins are flipped simultaneously, the probability of [#permalink]

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Probability of getting no heads = 1/64
Probability of getting at least 1 head = 63/64

Probability of getting no tails = 1/64
Probability of getting at least 1 tail = 63/64

Probability of getting at least 1 head and at least 1 tail = 63*63/64*64
63/64 is greater than 98%... So squaring that should be greater than 96%. Answer is E.
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Re: If six coins are flipped simultaneously, the probability of [#permalink]

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New post 24 Jan 2014, 08:49
Instead of assuming that \(\frac{31}{32}\) is 97%, you could do the following operation:

\(\frac{31}{32} = 1 - \frac{1}{32}\);

\(100 : 32 = 3,1...\) --> \(1 : 32 = 0,031...\)
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If six coins are flipped simultaneously, the probability of [#permalink]

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New post 12 Sep 2015, 22:37
hi all

can i use the combinations with repeating elements approach here?

clearly we meet the condition only if get either H H H H H H or T T T T T T out of a large number of possible combinations. If you have 30 secs left, play the game - choose between D and E.

1. As long as we have 6 elemens that can repeat: we have 6 * 6 * 6 * 6 * 6 * 6 total number of combinations.
2. HHHHHH in the combinatorics language means \(\frac{6!}{1!}\) = 6! = 720. TTTTTT is the same: 720. Sum up the 2: 1440
3. find the fraction via factorization: \(\frac{12*12*10}{3^6*2^6}\) = \(\frac{5}{162}\) which is obviosly is less than 6%=0.06
4. E is correct
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Re: If six coins are flipped simultaneously, the probability of [#permalink]

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New post 15 Feb 2016, 12:30
Total outcomes= 2*2*2*2*2*2 = 64
these outcomes wont work(where all heads or all tails)= 2
so not getting the outcome= \(\frac{2}{64}\)
getting the outcome = \(1 - \frac {2}{64}\) --> \(\frac{31}{32}\)

\(\frac{31}{32}\) = 0.96 =97%

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Re: If six coins are flipped simultaneously, the probability of [#permalink]

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Re: If six coins are flipped simultaneously, the probability of   [#permalink] 29 Oct 2017, 20:13
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