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If a fair two-sided coin is flipped 6 times, what is the probability

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If a fair two-sided coin is flipped 6 times, what is the probability  [#permalink]

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New post 06 Mar 2019, 14:05
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If a fair two-sided coin is flipped 6 times, what is the probability that tails is the result at least twice but at most 5 times?

A) \(\frac{5}{8}\)

B) \(\frac{3}{4}\)

C) \(\frac{7}{8}\)

D) \(\frac{57}{64}\)

E) \(\frac{15}{16}\)
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Re: If a fair two-sided coin is flipped 6 times, what is the probability  [#permalink]

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New post 07 Mar 2019, 09:30
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for getting at least two tails or at most 5 tails:
1) Probablity of getting 2 tails = 6C2/64 : (6C2 --> two tails can be from any of the 6 flips)
2) Probablity of getting 3 tails = 6C3/64
3) probablity of getting 4 tails = 6C4/64
4) probablity of getting 5 tails = 6C5/64

Probablity of at least 2 tails but at most 5 tails =(6C2+6C3+6C4+6C5)/64 = 56/64 = 7/8
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Re: If a fair two-sided coin is flipped 6 times, what is the probability  [#permalink]

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New post 10 Mar 2019, 19:45
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energetics wrote:
If a fair two-sided coin is flipped 6 times, what is the probability that tails is the result at least twice but at most 5 times?

A) \(\frac{5}{8}\)

B) \(\frac{3}{4}\)

C) \(\frac{7}{8}\)

D) \(\frac{57}{64}\)

E) \(\frac{15}{16}\)



We need to determine the following probabilities:

TTHHHH, TTTHHH, TTTTHH, and TTTTTH

P(TTHHHH) = (1/2)^6 = 1/64

TTHHHH can be arranged in 6C2 = 6!(2! x 4!) = (6x5)/2 = 15 ways, so the probability is 15/64.

P(TTTHHH) = (1/2)^6 = 1/64

TTTHHH can be arranged in 6C3 = 6!(3! x 3!) = (6x5x4)/3! = 20 ways, so the probability is 20/64.

P(TTTTHH) = (1/2)^6 = 1/64

TTTTHH can be arranged in 6C4 = 6!(2! x 4!) = (6x5)/2 = 15 ways, so the probability is 15/64.

P(TTTTTH) = (1/2)^6 = 1/64

TTTTTH can be arranged in 6C5 = 6!/5! = 6 ways, so the probability is 6/64.

Thus, the total probability is (15 + 20 + 15 + 6)/64 = 56/64 = 7/8.

Answer: C
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Re: If a fair two-sided coin is flipped 6 times, what is the probability  [#permalink]

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New post 06 Mar 2019, 21:38
The trick I am using to solve this question was taught to me by a channel on YouTube named Salman Gaffar. It’s Probability 07

So first let’s list down the possible outcomes

1) H H H H H H
2) H H H H H T
3) H H H H T T
4) H H H T T T
5) H H T T T T
6) H T T T T T
7) T T T T T T

We want to find out the probability of Atleast 2 tails and atmost 5 tails so option 1,2,7 are unfavourable and 3,4,5,6 are favourable

We can find probability of either one of them. Remember if we find probability of unfavourable outcomes we need to subtract it by 1 to get the probability of favourable outcomes. Let me show how

Let’s find the probability of option 1

1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 = 1/64

For option 2

1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 = 1/64 x 6!/5! = 6/64

For option 7

1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 = 1/64


So it’s either option 1 OR option 2 OR option 7 as unfavourable outcome

Since it’s OR we add, if its AND we multiply

1/64 + 1/64 + 6/64 = 8/64 or 1/8

1 - 1/8 = 7/8

Option C is the right answer

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If a fair two-sided coin is flipped 6 times, what is the probability  [#permalink]

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New post 21 Aug 2019, 04:47
energetics wrote:
If a fair two-sided coin is flipped 6 times, what is the probability that tails is the result at least twice but at most 5 times?

A) \(\frac{5}{8}\)
B) \(\frac{3}{4}\)
C) \(\frac{7}{8}\)
D) \(\frac{57}{64}\)
E) \(\frac{15}{16}\)


METHOD 1 (find the probability of each case and add them together)
TTHHHH: 1/64*(6!/2!4!)=1/64*15=15/64
TTTHHH: 1/64*(6!/3!3!)=1/64*20=20/64
TTTTHH: 1/64*(6!/4!2!)=1/64*15=15/64
TTTTTH: 1/64*(6!/5!1!)=1/64*6=6/64
total: 56/64=28/32=14/16=7/8

Answer (C)

METHOD 2 (find the probability of not getting the desired result and subtract 1 from the result)
HHHHHH [no tails]: 1/64*(6!/6!)=1/64*1=1/64
TTTTTTT [all tails]: 1/64*(6!/6!)=1/64*1=1/64
THHHHH [one tails]: 1/64*(6!/1!5!)=1/64*6=6/64
total: 1-(8/64)=56/64=28/32=14/16=7/8

Answer (C)
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Re: If a fair two-sided coin is flipped 6 times, what is the probability  [#permalink]

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New post 21 Aug 2019, 05:02
energetics wrote:
If a fair two-sided coin is flipped 6 times, what is the probability that tails is the result at least twice but at most 5 times?

A) \(\frac{5}{8}\)

B) \(\frac{3}{4}\)

C) \(\frac{7}{8}\)

D) \(\frac{57}{64}\)

E) \(\frac{15}{16}\)


The opposite even would be 0, 1, or 6 tails.

\(P(t = 0) = P(h = 6) = (\frac{1}{2})^6\)

\(P(t = 1) = \frac{6!}{5!}*(\frac{1}{2})*(\frac{1}{2})^5\). We are multiplying by 6!/5! because thhhhh can occur in 6!/5! = 6 different ways: thhhhh, hthhhh, hhthhh, hhhthh, hhhhth, hhhhht.

\(P(t = 6) = (\frac{1}{2})^6\)


\(P (t = 2, 3, 4, \ or \ 5) = 1- P(t = 0, 1, \ or \ 6) = 1 - ((\frac{1}{2})^6 + \frac{6!}{5!}*(\frac{1}{2})*(\frac{1}{2})^5 + (\frac{1}{2})^6) = \frac{7}{8}\)


Answer: C.
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Re: If a fair two-sided coin is flipped 6 times, what is the probability   [#permalink] 21 Aug 2019, 05:02
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