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

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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Welcome to Gmat Club forum.

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;

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.

Please ask if any question remains.

For more on probability and combinatorics please refer to the link: GMAT MATH BOOK
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the long way:
You have 4 options, tails twice, tails three times, tails 4 times, and tails 5 times

let's start with tails twice( this means 2 tails 4 heads)
6!/2!4! = 15

tails three times
6!/3!3! = 20

tails four times
6!/4!2! = 15

tails 5 times
6!/5!1! = 6

sum all and you get 56
56/64 = 7/8

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

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

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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What I meant was, when 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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Each coin can land on heads or tails, so 2 ways. We have 6 coins, so total number of outcomes is 2*2*2*2*2*2 = 2^6.
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Hi All,

In probability questions, there are two results that you can calculate - what you WANT to have happen or what you DON'T want to have happen. Since there are so many different ways to flip 2, 3, 4 or 5 tails, it will be easier for us to calculate what we DON'T want (0, 1 or 6 tails).

Since each toss has 2 possible outcomes (heads or tails), there are 2^6 = 64 different results for 6 coin flips.

Of those 64 options...

0 tails -->
HHHHHH = 1 option

1 tail -->
THHHHH
HTHHHH
HHTHHH
HHHTHH
HHHHTH
HHHHHT = 6 options

6 tails -->
TTTTTT = 1 option

1 + 6 + 1 = 8 options (of the 64) that we DON'T want...

Thus 64/64 - 8/64 = 56/64 = 7/8 that we DO want.

Final Answer:

GMAT assassins aren't born, they're made,
Rich

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

Posted from my mobile device

Bunnel and Iagomez, thanks for the timely responses!

Bunnel:
I was getting hung up on why 6C1 had to be multiplied by 6.


Thanks again to both of you.

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.

Hi Bunuel,

I understand the numerator part.
2C6 + 3C6 + 4C6 + 5C6 = 56

but how to calculate denominator part. I mean how can i count total no of combinations. I am not getting 64 .
Like in normal cases if we calculate for 6 ball, we take 6! as total no of combinations.
Please help

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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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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Approach 1

Given

• A fair 2 sided coin is flipped 6 times.


To Find

• The probability that tails will be the result at least twice, but not more than 5 times.


Approach and Working Out

• The tail should be at either 2, 3, 4, or 5.
• Total possible results = \(2^6\) = 64

o 2 Tails can appear in 6C2 ways = 15 ways
o 3 Tails can appear in 6C3 ways = 20 ways
o 4 Tails can appear in 6C4 ways = 15 ways
o 5 Tails can appear in 6C5 ways = 6 ways
• Answer = \(\frac{(15 + 20 + 15 + 6)}{64}\) =\(\frac{ 7}{8}\)

Correct Answer: Option C
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Approach 2

Given

• A fair 2 sided coin is flipped 6 times.

To Find

• The probability that tails will be the result at least twice, but not more than 5 times.

Approach and Working Out

• The tail should be at either 2, 3, 4, or 5.

o It means we should not count the ways when we get 0, 1, or 6 tails.

• Total possible results = 2^6 = 64

o 0 Tails can appear in 6C0 ways = 1 way
o 1 Tail can appear in 6C1 ways = 6 ways
o 6 Tails can appear in 6C6 ways = 1 ways
o Total ways not to get the desired number of tails = 1 + 6 + 1 = 8.
• Answer = 1 - \(\frac{8}{64}\) = \(\frac{7}{8}\)

Correct Answer: Option C
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The coin has a Head and a tail.

=> P(H) = P(T) = \(\frac{1}{2}\)

The probability that tails will be the result at least twice, but not more than 5 times: 1 - ['0' tails [All heads] + 1 tail + 6 tails]

'0' tail: \(\frac{1}{ (2^6)} = 1/ 64\)

'1' tails: \(^6{C_1} * \frac{1}{ (2^6)} = \frac{6}{64}\)

'6' tail: \(\frac{1}{ (2^6)} = 1/ 64\)

=> 1 - \([\frac{1 }{ 64} + \frac{6 }{ 64} + \frac{1 }{ 64}]\)

=> 1 - \(\frac{8}{64}\): 1 - \(\frac{1}{8}\) = \(\frac{7}{8}\)

Answer C

This can be solved using the binomial theorem easily.