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A fair 2-sided coin is flipped 6 times. What is the probability that

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A fair 2-sided coin is flipped 6 times. What is the probability that [#permalink]

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A fair 2-sided coin is flipped 6 times. What is the probability that tails will be the result at least twice, but not more than 5 times?

(A) 5/8
(B) 3/4
(C) 7/8
(D) 57/64
(E) 15/16
[Reveal] Spoiler: OA

Last edited by Bunuel on 16 Aug 2015, 14:39, edited 2 times in total.
Edited the question and added the OA

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Re: A fair 2-sided coin is flipped 6 times. What is the probability that [#permalink]

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melissawlim wrote:
A fair 2-sided coin is flipped 6 times. What is the probability that tails will be the result at least twice, but not more than 5 times?

(A) 5/8
(B) 3/4
(C) 7/8
(D) 57/64
(E) 15/16


It would be easier to calculate the probability of opposite event and subtract it from 1.
Opposite event: 0 tail, 1 tail, 6 tails.

Probability of getting no tails: \(\frac{1}{2^6}=\frac{1}{64}\);

Probability of getting 1 tail: \(6C1*\frac{1}{2^6}=\frac{6}{64}\), we must multiply by 6C1 or by 6 as tail can occur for any flip from 6, hence in 6 ways (*see more about this case below);

Probability of getting 6 tails: \(\frac{1}{2^6}=\frac{1}{64}\)

\(P=1-(\frac{1}{64}+\frac{6}{64}+\frac{1}{64})=\frac{56}{64}=\frac{7}{8}\)

Answer: C.


* Counting probability of getting 1 tail when flipped 6 times: 1 tail can occur in 6 different ways:

THHHHH
HTHHHH
HHTHHH
HHHTHH
HHHHTH
HHHHHT

Generally probability of occurring event k times in n-time sequence could be expressed as:

\(P = C^n_k*p^k*(1-p)^{n-k}\)

In our case \(k=1\) and \(n=6\), so we get:

\(P = C^6_1*\frac{1}{2}*\frac{1}{2^5}=6*\frac{1}{2^6}=\frac{6}{64}\)

So \(\frac{1}{64}\) should be multiplied by \(C^6_1\), which is \(6\).
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Re: A fair 2-sided coin is flipped 6 times. What is the probability that [#permalink]

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lets find it out the other way
P of all head(or no tail)=1/2^6=1/64
P of all tail (or no head)=1/2^6=1/64
P of one tail=1/2^6 * 6 { multiply by 6 as there are 6 ways we can get one tail and each is having a probability of 1/2}
tot=1/64 + 1/64 + 6/64=1/8
reqd P=1-tot=1-1/8=7/8

C
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Re: A fair 2-sided coin is flipped 6 times. What is the probability that [#permalink]

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New post 19 Jun 2014, 19:11
Hi Bunuel,

Shouldn't the value of K be equal to 2 as we are looking for atleast twice.And secondly can you please explain me how do we restrict here that tails do not occur for more than five times by using the formula.

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Re: A fair 2-sided coin is flipped 6 times. What is the probability that [#permalink]

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maggie27 wrote:
Hi Bunuel,

Shouldn't the value of K be equal to 2 as we are looking for atleast twice.And secondly can you please explain me how do we restrict here that tails do not occur for more than five times by using the formula.


The example there with formula is for getting 1 tail when flipped 6 times.

The solution uses P(good) = 1 - P(bad). So, P(2, 3, 4, or 5 tails) = 1 - P(0, 1, or 6 tails.)
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Re: A fair 2-sided coin is flipped 6 times. What is the probability that [#permalink]

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New post 20 Jun 2014, 06:29
My bad :(
Thank You Bunuel :)

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Re: A fair 2-sided coin is flipped 6 times. What is the probability that [#permalink]

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melissawlim wrote:
A fair 2-sided coin is flipped 6 times. What is the probability that tails will be the result at least twice, but not more than 5 times?

(A) 5/8
(B) 3/4
(C) 7/8
(D) 57/64
(E) 15/16


Total possible outcomes when coined is tossed 6 time=2^6=4*4*4=64
Total possible outcomes getting 2 or 3 or 4 or 5 tails= 6C2+6C3+6C4+6C5=(6*5)/2+(6*5*4)/(3*2)+(6*5)/2+6=15+20+15+6=56
Probability of getting atleast 2 but not more than 5 times tails=56/64=7/8

Ans=C

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Re: A fair 2-sided coin is flipped 6 times. What is the probability that [#permalink]

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New post 20 Jun 2014, 22:35
Forgot considering 0 events))

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Re: A fair 2-sided coin is flipped 6 times. What is the probability that [#permalink]

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New post 05 Nov 2017, 07:32
melissawlim wrote:
A fair 2-sided coin is flipped 6 times. What is the probability that tails will be the result at least twice, but not more than 5 times?

(A) 5/8
(B) 3/4
(C) 7/8
(D) 57/64
(E) 15/16


We need to determine the probability of flipping tails 2, 3, 4, or 5 times in 6 flips. Thus, we could use the following formula:

P(flipping tails 2, 3, 4, or 5 times in 6 flips) = 1 - P(selecting tails 0, 1, or 6 times)

P(tails zero times) can be denoted and calculated as:

H-H-H-H-H-H = (1/2)^6 = 1/64

P(tails 1 time) can be denoted and calculated as:

T-H-H-H-H-H = (1/2)^6 = 1/64

However, T-H-H-H-H-H can be arranged in 6!/5! = 6 ways, so the overall probability is 1/64 x 6 = 6/64.

P(tails 6 times) can be denoted and calculated as:

T-T-T-T-T-T = (1/2)^6 = 1/64

Thus:

P(flipping tails 2, 3, 4, or 5 times in 6 flips) = 1 - (1/64 + 6/64 + 1/64) = 1 - 8/64 = 1- 1/8 = 7/8.

Alternate Solution:

Let x = the number of times tails will appear. Thus, we have:

P(2 ≤ x ≤ 5) = 1 - P(x = 0) - P(x = 1) - P(x = 6)

Let’s determine P(x = 0), P(x = 1), and P(x = 6):

P(x = 0) = (½)^6 = 1/64 (Note: 0 tails means: HHHHHH)

P(x = 1) = (½)^6 x 6 = 6/64 (Note: 1 tails means: THHHHH and 5 other “1T and 5H” arrangements)

P(x = 6) = (½)^6 = 1/64 (Note: 6 tails mean: TTTTTT)

Thus:

P(2 ≤ x ≤ 5) = 1 -1/64 - 6/64 - 1/64 = 56/64 = ⅞

Answer: C
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Re: A fair 2-sided coin is flipped 6 times. What is the probability that   [#permalink] 05 Nov 2017, 07:32
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